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Chemistry Slime Tutorial (19-24) Flipbook PDF
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WHAT’S AHEAD 19.1 SPONTANEOUS PROCESSES
19.3 MOLECULAR INTERPRETATION OF ENTROPY
We see that changes that occur in nature have a directional character. They move spontaneously in one direction but not in the reverse direction.
On the molecular level, the entropy of a system is related to the number of accessible microstates. The entropy of the system increases as the randomness of the system increases. The third law of thermodynamics states that, at 0 K, the entropy of a perfect crystalline solid is zero.
19.2 ENTROPY AND THE SECOND LAW OF THERMODYNAMICS We discuss entropy, a thermodynamic state function that is important in determining whether a process is spontaneous. The second law of thermodynamics tells us that in any spontaneous process the entropy of the universe (system plus surroundings) increases.
THE SKYLINE OF HONG KONG The construction of our human environment entails the use of enormous amounts of energy to create complex ordered structures, such as modern skyscrapers.
19.4 ENTROPY CHANGES IN CHEMICAL REACTIONS Using tabulated standard molar entropies, we can calculate the standard entropy changes for systems undergoing reaction.
19
19.5 GIBBS FREE ENERGY
19.7 FREE ENERGY AND THE EQUILIBRIUM CONSTANT
We encounter another thermodynamic state function, free energy (or Gibbs free energy), a measure of how far removed a system is from equilibrium. The change in free energy measures the maximum amount of useful work obtainable from a process and tells us the direction in which a chemical reaction is spontaneous.
Finally, we consider how the standard free-energy change for a chemical reaction can be used to calculate the equilibrium constant for the reaction.
19.6 FREE ENERGY AND TEMPERATURE We consider how the relationship among free-energy change, enthalpy change, and entropy change provides insight into how temperature affects the spontaneity of a process.
CHEMICAL THERMODYNAMICS Earth’s resources to create impressive, highly ordered structures, such as the beautiful skyline in the chapter-opening photograph. Our modern society depends heavily on the design of chemical reactions that produce specific useful substances from natural and synthetic materials. HUMANKIND HAS LEARNED TO HARNESS
Two of the most important questions chemists ask when designing and using chemical reactions are “How fast is the reaction?” and “How far does it proceed?” The first question is addressed by chemical kinetics, which we discussed in Chapter 14. The second question involves the equilibrium constant, the focus of Chapter 15. Let’s briefly review how these concepts are related. In Chapter 14 we learned that the rate of any chemical reaction is controlled largely by a factor related to energy, namely, the activation energy of the reaction. •(Section 14.5) In general, the lower the activation energy, the faster a reaction proceeds. In Chapter 15 we saw that chemical equilibrium is reached when a given reaction and its reverse reaction occur at the same rate. •(Section 15.1) Because reaction rates are closely tied to energy, it is logical that equilibrium also depends in some way on energy. In this chapter we explore the connection between energy and the extent of a reaction. Doing so requires a deeper look at chemical thermodynamics, the area of chemistry that deals with energy relationships. We first encountered thermodynamics in Chapter 5, where we discussed the nature of energy, 785
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the first law of thermodynamics, and the concept of enthalpy. Recall that the enthalpy change for any system is the heat transferred between the system and its surroundings during a constant-pressure process. •(Section 5.3) In the “Strategies in Chemistry” box in Section 5.4, we pointed out that the enthalpy change that takes place during a reaction is an important guide as to whether the reaction is likely to proceed. Now we will see that reactions involve not only changes in enthalpy but also changes in entropy—another important thermodynamic quantity. Our discussion of entropy will lead us to the second law of thermodynamics, which provides insight into why physical and chemical changes tend to favor one direction over another. We drop a brick, for example, and it falls to the ground. We do not expect the brick to spontaneously rise from the ground to our outstretched hand. We light a candle, and it burns down. We do not expect a half-consumed candle to regenerate itself spontaneously, even if we have captured all the gases produced when the candle burned. Thermodynamics helps us understand the significance of this directional character of processes, regardless of whether they are exothermic or endothermic.
19.1 | SPONTANEOUS PROCESSES GO FIGURE
Does the potential energy of the eggs change during this process?
Spontaneous
Not spontaneous
쑿 FIGURE 19.1 A spontaneous process!
The first law of thermodynamics states that energy is conserved. •(Section 5.2) In other words, energy is neither created nor destroyed in any process, whether that process is a brick falling, a candle burning, or an ice cube melting. Energy can be transferred between a system and the surroundings and can be converted from one form to another, but the total energy of the universe remains constant. We expressed this law mathematically as ¢E = q + w, where ¢E is the change in the internal energy of a system, q is the heat absorbed (or released) by the system from (or to) the surroundings, and w is the work done on the system by the surroundings, or on the surroundings by the system. Remember that q 7 0 means that the system is absorbing heat from the surroundings, and w 7 0 means that the surroundings are doing work on the system. The first law helps us balance the books, so to speak, on the heat transferred between a system and its surroundings and the work done by or on a system. However, because energy is conserved, we can’t simply use the value of ¢E to tell us whether a process is favored to occur because anything we do to lower the energy of the system raises the energy of the surroundings, and vice versa. Nevertheless, experience tells us that certain processes always occur, even though the energy of the universe is conserved. Water placed in a freezer turns into ice, for instance, and if you touch a hot object, heat is transferred to your hand. The first law guarantees that energy is conserved in these processes, and yet they occur without any outside intervention. We say they are spontaneous. A spontaneous process is one that proceeds on its own without any outside assistance. A spontaneous process occurs in one direction only, and the reverse of any spontaneous process is always nonspontaneous. Drop an egg above a hard surface, for example, and it breaks on impact (씱 FIGURE 19.1). Now, imagine seeing a video clip in which a broken egg rises from the floor, reassembles itself, and ends up in someone’s hand. You would conclude that the video is running in reverse because you know that broken eggs simply do not magically rise and reassemble themselves! An egg falling and breaking is spontaneous. The reverse process is nonspontaneous, even though energy is conserved in both processes. We know other spontaneous and nonspontaneous processes that relate more directly to our study of chemistry. For example, a gas spontaneously expands into a vacuum (씰 FIGURE 19.2), but the reverse process, in which the gas moves back entirely into one of the flasks, does not happen. In other words, expansion of the gas is spontaneous, but the reverse process is nonspontaneous. In general, processes that are spontaneous in one direction are nonspontaneous in the opposite direction. Experimental conditions, such as temperature and pressure, are often important in determining whether a process is spontaneous. We are all familiar with situations in which a forward process is spontaneous at one temperature but the reverse process is
SECTION 19.1
spontaneous at a different temperature. Consider, for example, ice melting. At atmospheric pressure, when the temperature of the surroundings is above 0 °C, ice melts spontaneously, and the reverse process— liquid water turning into ice—is not spontaneous. However, when the temperature of the surroundings is below 0 °C, the opposite is true—liquid water turns to ice spontaneously, but the reverse process is not spontaneous (쑼 FIGURE 19.3). What happens at T = 0 °C, the normal melting point of water, when the flask of Figure 19.3 contains both water and ice? At the normal melting point of a substance, the solid and liquid phases are in equilibrium. •(Section 11.6) At this temperature, the two phases are interconverting at the same rate and there is no preferred direction for the process. It is important to realize that the fact that a process is spontaneous does not necessarily mean that it will occur at an observable rate. A chemical reaction is spontaneous if it occurs on its own accord, regardless of its speed. A spontaneous reaction can be very fast, as in the case of acid–base neutralization, or very slow, as in the rusting of iron. Thermodynamics tells us the direction and extent of a reaction but nothing about the speed.
Spontaneous Processes
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GO FIGURE
If flask B were smaller than flask A, would the final pressure after the stopcock is opened be greater than, equal to, or less than 0.5 atm? Closed stopcock A
B
Gas at 1 atm
Evacuated flask 0 atm
When stopcock opens, gas expands to occupy both flasks
GIVE IT SOME THOUGHT If a process is nonspontaneous, does that mean the process cannot occur under any circumstances?
A
B
0.5 atm
0.5 atm
GO FIGURE
In which direction is this process exothermic?
Spontaneous for T 0 C
This process is spontaneous
All gas molecules move back into flask A
A
B
1 atm
0 atm
Spontaneous for T 0 C
This process is not spontaneous 쑿 FIGURE 19.3 Spontaneity can depend on temperature. At T 7 0 °C, ice melts spontaneously to liquid water. At T 6 0 °C, the reverse process, water freezing to ice, is spontaneous. At T = 0 °C the two states are in equilibrium.
SAMPLE EXERCISE 19.1
Identifying Spontaneous Processes
Predict whether each process is spontaneous as described, spontaneous in the reverse direction, or in equilibrium: (a) Water at 40 °C gets hotter when a piece of metal heated to 150 °C is added. (b) Water at room temperature decomposes into H2(g) and O2(g). (c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1 °C. SOLUTION Analyze We are asked to judge whether each process is spontaneous in the direction indicated, in the reverse direction, or in neither direction. Plan We need to think about whether each process is consistent with our experience about the natural direction of events or whether we expect the reverse process to occur. Solve (a) This process is spontaneous. Whenever two objects at different temperatures are brought into contact, heat is transferred from the hotter object to the colder one. • (Section 5.1) Thus, heat is transferred from the hot metal to the cooler water. The final temperature, after the metal and water achieve the same temperature (thermal equilibrium), will be somewhere
쑿 FIGURE 19.2 Expansion of a gas into an evacuated space is a spontaneous process. The reverse process—gas molecules initially distributed evenly in two flasks all moving into one flask—is not spontaneous.
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between the initial temperatures of the metal and the water. (b) Experience tells us that this process is not spontaneous—we certainly have never seen hydrogen and oxygen gases spontaneously bubbling up out of water! Rather, the reverse process—the reaction of H2 and O2 to form H2O—is spontaneous. (c) The normal boiling point is the temperature at which a vapor at 1 atm is in equilibrium with its liquid. Thus, this is an equilibrium situation. If the temperature were below 80.1 °C, condensation would be spontaneous. PRACTICE EXERCISE At 1 atm pressure, CO2(s) sublimes at -78 °C. Is this process spontaneous at -100 °C and 1 atm pressure? Answer: No, the reverse process is spontaneous at this temperature.
Seeking a Criterion for Spontaneity A marble rolling down an incline or a brick falling from your hand loses potential energy. The loss of some form of energy is a common feature of spontaneous change in mechanical systems. During the 1870s Marcellin Bertholet (1827–1907), a famous chemist of that era, suggested that the direction of spontaneous changes in chemical systems is determined by the loss of energy. He proposed that all spontaneous chemical and physical changes are exothermic. It takes only a few moments, however, to find exceptions to this generalization. For example, the melting of ice at room temperature is spontaneous and endothermic. Similarly, many spontaneous dissolution processes, such as the dissolving of NH4NO3, are endothermic, as we discovered in Section 13.1. We conclude that although the majority of spontaneous reactions are exothermic, there are spontaneous endothermic ones as well. Clearly, some other factor must be at work in determining the natural direction of processes. To understand why certain processes are spontaneous, we need to consider more closely the ways in which the state of a system can change. Recall from Section 5.2 that quantities such as temperature, internal energy, and enthalpy are state functions, properties that define a state and do not depend on how we reach that state. The heat transferred between a system and its surroundings, q, and the work done by or on the system, w, are not state functions—their values depend on the specific path taken between states. One key to understanding spontaneity is understanding differences in the paths between states.
Reversible and Irreversible Processes In 1824 a 28-year-old French engineer named Sadi Carnot (1796–1832) published an analysis of the factors that determine how efficiently a steam engine can convert heat to work. Carnot considered what an ideal engine, one with the highest possible efficiency, would be like. He observed that it is impossible to convert the energy content of a fuel completely to work because a significant amount of heat is always lost to the surroundings. Carnot’s analysis gave insight into how to build better, more efficient engines, and it was one of the earliest studies in what has developed into the discipline of thermodynamics. An ideal engine operates under an ideal set of conditions in which all the processes are reversible. A reversible process is a specific way in which a system changes its state. In a reversible process, the change occurs in such a way that the system and surroundings can be restored to their original states by exactly reversing the change. In other words, we can restore the system to its original condition with no net change to either the system or its surroundings. An irreversible process is one that cannot simply be reversed to restore the system and its surroundings to their original states. What Carnot discovered is that the amount of work we can extract from any process depends on the manner in which the process is carried out. He concluded that a reversible change produces the maximum amount of work that can be done by a system on its surroundings.
SECTION 19.1
Spontaneous Processes
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GO FIGURE
If the flow of heat into or out of the system is to be reversible, what must be true of D T? Small increment of heat transferred from system to surroundings
System at higher temperature T + δT
Small increment of heat transferred to system from surroundings
System at lower temperature T – δT
System
System
Surroundings
Surroundings
Surroundings at temperature T
Surroundings at temperature T
(a)
(b)
씱 FIGURE 19.4 Reversible flow of heat. Heat can flow reversibly between a system and its surroundings only if the two have an infinitesimally small difference in temperature dT. (a) Increasing the temperature of the system by dT causes heat to flow from the hotter system to the colder surroundings. (b) Decreasing the temperature of the system by dT causes heat to flow from the hotter surroundings to the colder system.
GIVE IT SOME THOUGHT Suppose you have a system made up of water only, with the container and everything beyond being the surroundings. Consider a process in which the water is first evaporated and then condensed back into its original container. Is this two-step process necessarily reversible?
Let’s next examine some aspects of reversible and irreversible processes, first with respect to the transfer of heat. When two objects at different temperatures are in contact, heat flows spontaneously from the hotter object to the colder one. Because it is impossible to make heat flow in the opposite direction, from colder object to hotter one, the flow of heat is an irreversible process. Given these facts, can we imagine any conditions under which heat transfer can be made reversible? To answer this question, we must consider temperature differences that are infinitesimally small, as opposed to the discrete temperature differences with which we are most familiar. For example, consider a system and its surroundings at essentially the same temperature, with just an infinitesimal temperature difference dT between them (쑿 FIGURE 19.4). If the surroundings are at temperature T and the system is at the infinitesimally higher temperature T + dT, then an infinitesimal amount of heat flows from system to surroundings. We can reverse the direction of heat flow by making an infinitesimal change of temperature in the opposite direction, lowering the system temperature to T - dT. Now the direction of heat flow is from surroundings to system. Reversible processes are those that reverse direction whenever an infinitesimal change is made in some property of the system.* Now let’s consider another example, the expansion of an ideal gas at constant temperature (referred to as an isothermal process). In the cylinder-piston arrangement of 쑼 FIGURE 19.5, when the partition is removed, the gas expands spontaneously to fill the Lift partition Piston
Do work on piston, compress gas
Movable partition Work
Vacuum
Gas Irreversible expansion of gas (work done by system 0)
Compression (work done on system 0)
*For a process to be truly reversible, the amounts of heat must be infinitesimally small and the transfer of heat must occur infinitely slowly; thus, no process that we can observe is truly reversible. The notion of infinitesimal amounts are related to the infinitesimals that you may have studied in a calculus course.
씱 FIGURE 19.5 An irreversible process. Initially an ideal gas is confined to the right half of a cylinder. When the partition is removed, the gas spontaneously expands to fill the whole cylinder. No work is done by the system during this expansion. Using the piston to compress the gas back to its original state requires the surroundings to do work on the system.
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evacuated space. Can we determine whether this particular isothermal expansion is reversible or irreversible? Because the gas expands into a vacuum with no external pressure, it does no P-V work on the surroundings. •(Section 5.3) Thus, for the expansion, w = 0. We can use the piston to compress the gas back to its original state, but doing so requires that the surroundings do work on the system, meaning that w 7 0 for the compression. In other words, the path that restores the system to its original state requires a different value of w (and, by the first law, a different value of q) than the path by which the system was first changed. The fact that the same path can’t be followed to restore the system to its original state indicates that the process is irreversible. What might a reversible isothermal expansion of an ideal gas be? This process will occur only if initially, when the gas is confined to half the cylinder, the external pressure acting on the piston exactly balances the pressure exerted by the gas on the piston. If the external pressure is reduced infinitely slowly, the piston will move outward, allowing the pressure of the confined gas to readjust to maintain the pressure balance. This infinitely slow process in which the external pressure and internal pressure are always in equilibrium is reversible. If we reverse the process and compress the gas in the same infinitely slow manner, we can return the gas to its original volume. The complete cycle of expansion and compression in this hypothetical process, moreover, is accomplished without any net change to the surroundings. Because real processes can at best only approximate the infinitely slow change associated with reversible processes, all real processes are irreversible. Further, as noted earlier in this discussion, the reverse of any spontaneous process is a nonspontaneous process. A nonspontaneous process can occur only if the surroundings do work on the system. Thus, any spontaneous process is irreversible. Even if we return the system to the original condition, the surroundings will have changed.
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19.2 ENTROPY AND THE SECOND LAW OF THERMODYNAMICS How can we use the fact that any spontaneous process is irreversible to make predictions about the spontaneity of an unfamiliar process? Understanding spontaneity requires us to examine the thermodynamic quantity called entropy, which was first mentioned in Section 13.1. In general, entropy is associated either with the extent of randomness in a system or with the extent to which energy is distributed among the various motions of the molecules of the system. In this section we consider how we can relate entropy changes to heat transfer and temperature. Our analysis will bring us to a profound statement about spontaneity that we call the second law of thermodynamics.
Entropy Change The entropy, S, of a system is a state function just like internal energy, E, and enthalpy, H. As with these other quantities, the value of S is a characteristic of the state of a system. •(Section 5.2) Thus, the change in entropy, ¢S, in a system depends only on the initial and final states of the system and not on the path taken from one state to the other: ¢S = Sfinal - Sinitial
[19.1]
For the special case of an isothermal process, ¢S is equal to the heat that would be transferred if the process were reversible, qrev, divided by the absolute temperature at which the process occurs: ¢S =
qrev T
1constant T2
[19.2]
Although there are many possible paths that can take the system from one state to another, only one path is associated with a reversible process. Thus, the value of qrev is uniquely defined for any two states of the system. Because S is a state function, we can use Equation 19.2 to calculate ¢S for any isothermal process between states, not just the reversible one.
SECTION 19.2
Entropy and the Second Law of Thermodynamics
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GIVE IT SOME THOUGHT How do we reconcile the fact that S is a state function but ¢S depends on q, which is not a state function?
¢S for Phase Changes The melting of a substance at its melting point and the vaporization of a substance at its boiling point are isothermal processes. • (Section 11.4) Consider the melting of ice. At 1 atm pressure, ice and liquid water are in equilibrium at 0 °C. Imagine melting 1 mol of ice at 0 °C, 1 atm to form 1 mol of liquid water at 0 °C, 1 atm. We can achieve this change by adding a certain amount of heat to the system from the surroundings: q = ¢Hfusion. Now imagine that we add the heat infinitely slowly, raising the temperature of the surroundings only infinitesimally above 0 °C. When we make the change in this fashion, the process is reversible because we can reverse it by infinitely slowly removing the same amount of heat, ¢Hfusion, from the system, using immediate surroundings that are infinitesimally below 0 °C. Thus, qrev = ¢Hfusion for the melting of ice at T = 0 °C = 273 K. The enthalpy of fusion for H2O is ¢Hfusion = 6.01 kJ>mol (a positive value because melting is an endothermic process). Thus, we can use Equation 19.2 to calculate ¢Sfusion for melting 1 mol of ice at 273 K: ¢S fusion
(1 mol2(6.01 * 103 J>mol2 qrev ¢Hfusion = = = = 22.0 J>K T T 273 K
Notice (a) that we must use the absolute temperature in Equation 19.2 and (b) that the units for ¢S, J>K, are energy divided by absolute temperature, as we expect from Equation 19.2.
SAMPLE EXERCISE 19.2
Calculating ¢S for a Phase Change
Elemental mercury is a silver liquid at room temperature. Its normal freezing point is -38.9 °C, and its molar enthalpy of fusion is ¢Hfusion = 2.29 kJ>mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? SOLUTION Analyze We first recognize that freezing is an exothermic process, which means heat is transferred from system to surroundings and q 6 0. The enthalpy of fusion refers to the process of melting. Because freezing is the reverse of melting, the enthalpy change that accompanies the freezing of 1 mol of Hg is - ¢Hfusion = -2.29 kJ>mol.
Plan We can use - ¢Hfusion and the atomic weight of Hg to calculate q for freezing 50.0 g of Hg. Then we use this value of q as qrev in Equation 19.2 to determine ¢S for the system.
Solve For q we have
q = 150.0 g Hg2a
Before using Equation 19.2, we must first convert the given Celsius temperature to kelvins:
-38.9 °C = 1-38.9 + 273.152 K = 234.3 K
We can now calculate ¢Ssys:
¢Ssys =
1 mol Hg -2.29 kJ 1000 J ba ba b = -571 200.59 g Hg 1 mol Hg 1 kJ
qrev -571 J = = -2.44 J>K T 234.3 K
Check The entropy change is negative because our qrev value is negative, which it must be because heat flows out of the system in this exothermic process. Comment This procedure can be used to calculate ¢S for other isothermal phase changes, such as the vaporization of a liquid at its boiling point. PRACTICE EXERCISE The normal boiling point of ethanol, C2H5OH, is 78.3 °C, and its molar enthalpy of vaporization is 38.56 kJ>mol. What is the change in entropy in the system when 68.3 g of C2H5OH(g) at 1 atm condenses to liquid at the normal boiling point? Answer: -163 J>K
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A CLOSER LOOK THE ENTROPY CHANGE WHEN A GAS EXPANDS ISOTHERMALLY In general, the entropy of any system increases as the system becomes more random or more spread out. Thus, we expect the spontaneous expansion of a gas to result in an increase in entropy. To see how this entropy increase can be calculated, consider the expansion of an ideal gas that is initially constrained by a piston, as in the rightmost part of Figure 19.5. Imagine that we allow the gas to undergo a reversible isothermal expansion by infinitesimally decreasing the external pressure on the piston. The work done on the surroundings by the reversible expansion of the system against the piston can be calculated with the aid of calculus (we do not show the derivation): wrev = -nRT ln
V2 V1
In this equation, n is the number of moles of gas, R is the gas constant • (Section 10.4), T is the absolute temperature, V1 is the initial volume, and V2 is the final volume. Notice that if V2 7 V1, as it must be in our expansion, then wrev 6 0, meaning that the expanding gas does work on the surroundings. One characteristic of an ideal gas is that its internal energy depends only on temperature, not on pressure. Thus, when an ideal
gas expands isothermally, ¢E = 0. Because ¢E = qrev + wrev = 0, we see that qrev = -wrev = nRT ln(V2>V12. Then, using Equation 19.2, we can calculate the entropy change in the system: V2 nRT ln qrev V1 V2 [19.3] = = nR ln ¢Ssys = T T V1 From the ideal-gas equation, we can calculate the number of moles in 1.00 L of an ideal gas at 1.00 atm and 0 °C by using the value 0.08206 L-atm>mol-K for R: n =
11.00 atm211.00 L2 PV = = 4.46 * 10-2 mol RT 10.08206 L-atm>mol-K21273 K2
The gas constant, R, can also be expressed as 8.314 J>mol-K (Table 10.2), and this is the value we must use in Equation 19.3 because we want our answer to be expressed in terms of J rather than in L-atm. Thus, for the expansion of the gas from 1.00 L to 2.00 L, we have ¢Ssys = 14.46 * 10-2 mol2a8.314
J 2.00 L b aln b mol-K 1.00 L
= 0.26 J>K In Section 19.3 we will see that this increase in entropy is a measure of the increased randomness of the molecules because of the expansion. RELATED EXERCISES: 19.29, 19.30, and 19.106
The Second Law of Thermodynamics The key idea of the first law of thermodynamics is that energy is conserved in any process. •(Section 5.2) Entropy, however, is not conserved. For any spontaneous process, the total change in entropy, which is the sum of the entropy change of the system plus the entropy change of the surroundings, is greater than zero. Let’s illustrate this generalization by calculating the entropy change of a system and the entropy change of its surroundings when our system is 1 mol of ice (a piece roughly the size of an ice cube) melting in the palm of your hand, which is part of the surroundings. The process is not reversible because the system and surroundings are at different temperatures. Nevertheless, because ¢S is a state function, its value is the same regardless of whether the process is reversible or irreversible. We calculated the entropy change of the system just before Sample Exercise 19.2: 11 mol216.01 * 103 J>mol2 qrev ¢Ssys = = = 22.0 J>K T 273 K The surroundings immediately in contact with the ice are your hand, which we assume is at body temperature, 37 °C = 310 K. The quantity of heat lost by your hand is -6.01 * 103 J>mol, which is equal in magnitude to the quantity of heat gained by the ice but has the opposite sign. Hence, the entropy change of the surroundings is 11 mol21-6.01 * 103 J>mol2 qrev ¢Ssurr = = = -19.4 J>K T 310 K Thus, the total entropy change is positive: ¢Stotal = ¢Ssys + ¢Ssurr = 122.0 J>K2 + 1-19.4 J>K2 = 2.6 J>K If the temperature of the surroundings were not 310 K but rather some temperature infinitesimally above 273 K, the melting would be reversible instead of irreversible. In that case the entropy change of the surroundings would equal -22.0 J>K and ¢Stotal would be zero.
SECTION 19.3
Molecular Interpretation of Entropy
In general, any irreversible process results in an increase in total entropy, whereas any reversible process results in no overall change in entropy. This statement is known as the second law of thermodynamics. The sum of the entropy of a system plus the entropy of the surroundings is everything there is, and so we refer to the total entropy change as the entropy change of the universe, ¢Suniv. We can therefore state the second law of thermodynamics in terms of two equations: Reversible Process: ¢Suniv = ¢Ssys + ¢Ssurr = 0 Irreversible Process: ¢Suniv = ¢Ssys + ¢Ssurr 7 0
[19.4]
Because spontaneous processes are irreversible, we can say that the entropy of the universe increases in any spontaneous process. This profound generalization is yet another way of expressing the second law of thermodynamics. GIVE IT SOME THOUGHT The rusting of iron is spontaneous and is accompanied by a decrease in the entropy of the system (the iron and oxygen). What can we conclude about the entropy change of the surroundings?
The second law of thermodynamics tells us the essential character of any spontaneous change—it is always accompanied by an increase in the entropy of the universe. We can use this criterion to predict whether a given process is spontaneous or not. Before seeing how this is done, however, we will find it useful to explore entropy from a molecular perspective. A word on notation before we proceed. Throughout most of the remainder of this chapter, we will focus on systems rather than surroundings. To simplify the notation, we will usually refer to the entropy change of the system as ¢S rather than explicitly indicating ¢Ssys.
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19.3 MOLECULAR INTERPRETATION OF ENTROPY As chemists, we are interested in molecules. What does entropy have to do with them and with their transformations? What molecular property does entropy reflect? Ludwig Boltzmann (1844–1906) gave conceptual meaning to the notion of entropy, and to understand his contribution, we need to examine the ways in which we can interpret entropy at the molecular level.
Expansion of a Gas at the Molecular Level In discussing Figure 19.2, we talked about the expansion of a gas into a vacuum as a spontaneous process. We now understand that it is an irreversible process and that the entropy of the universe increases during the expansion. How can we explain the spontaneity of this process at the molecular level? We can get a sense of what makes this expansion spontaneous by envisioning the gas as a collection of particles in constant motion, as we did in discussing the kinetic-molecular theory of gases. •(Section 10.7) When the stopcock in Figure 19.2 is opened, we can view the expansion of the gas as the ultimate result of the gas molecules moving randomly throughout the larger volume. Let’s look at this idea more closely by tracking two of the gas molecules as they move around. Before the stopcock is opened, both molecules are confined to the left flask, as shown in 씰 FIGURE 19.6(a). After the stopcock is opened, the molecules travel randomly throughout the entire apparatus. As Figure 19.6(b) shows, there are four possible arrangements for the two molecules once both flasks are available to them. Because the molecular motion is random, all four arrangements are equally likely. Note that now only one arrangement corresponds to the situation before the stopcock was opened: both molecules in the left flask.
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(a) The two molecules are colored red and blue to keep track of them.
(b) Four possible arrangements (microstates) once the stopcock is opened.
쑿 FIGURE 19.6 Possible arrangements of two gas molecules in two flasks. (a) Before the stopcock is opened, both molecules are in the left flask. (b) After the stopcock is opened, there are four possible arrangements of the two molecules.
Figure 19.6(b) shows that with both flasks available to the molecules, the probability of the red molecule being in the left flask is two in four (top right and bottom left arrangements), and the probability of the blue molecule being in the left flask is the same (top left and bottom left arrangements). Because the probability is 2>4 = 1>2 that each molecule is in the left flask, the probability that both are there is 11> 222 = 1> 4. If we apply the same analysis to three gas molecules, we find that the probability that all three are in the left flask at the same time is 11> 223 = 1>8. Now let’s consider a mole of gas. The probability that all the molecules are in the left flask at the same time is (1> 2)N, where N = 6.02 * 1023. This is a vanishingly small number! Thus, there is essentially zero likelihood that all the gas molecules will be in the left flask at the same time. This analysis of the microscopic behavior of the gas molecules leads to the expected macroscopic behavior: The gas spontaneously expands to fill both the left and right flasks, and it does not spontaneously all go back in the left flask. This molecular view of gas expansion shows the tendency of the molecules to “spread out” among the different arrangements they can take. Before the stopcock is opened, there is only one possible arrangement: all molecules in the left flask. When the stopcock is opened, the arrangement in which all the molecules are in the left flask is but one of an extremely large number of possible arrangements. The most probable arrangements by far are those in which there are essentially equal numbers of molecules in the two flasks. When the gas spreads throughout the apparatus, any given molecule can be in either flask rather than confined to the left flask. We say that with the stopcock opened, the arrangement of gas molecules is more random or disordered than when the molecules are all confined in the left flask. We will see this notion of increasing randomness helps us understand entropy at the molecular level.
Boltzmann’s Equation and Microstates The science of thermodynamics developed as a means of describing the properties of matter in our macroscopic world without regard to microscopic structure. In fact, thermodynamics was a well-developed field before the modern view of atomic and molecular structure was even known. The thermodynamic properties of water, for example, addressed the behavior of bulk water (or ice or water vapor) as a substance without considering any specific properties of individual H2O molecules. To connect the microscopic and macroscopic descriptions of matter, scientists have developed the field of statistical thermodynamics, which uses the tools of statistics and probability to link the microscopic and macroscopic worlds. Here we show how entropy, which is a property of bulk matter, can be connected to the behavior of atoms and molecules. Because the mathematics of statistical thermodynamics is complex, our discussion will be largely conceptual. In our discussion of two gas molecules in the two-flask system in Figure 19.6, we saw that the number of possible arrangements helped explain why the gas expands.
SECTION 19.3
Molecular Interpretation of Entropy
795
Suppose we now consider one mole of an ideal gas in a particular thermodynamic state, which we can define by specifying the temperature, T, and volume, V, of the gas. What is happening to this gas at the microscopic level, and how does what is going on at the microscopic level relate to the entropy of the gas? Imagine taking a snapshot of the positions and speeds of all the molecules at a given instant. The speed of each molecule tells us its kinetic energy. That particular set of 6 * 1023 positions and kinetic energies of the individual gas molecules is what we call a microstate of the system. A microstate is a single possible arrangement of the positions and kinetic energies of the gas molecules when the gas is in a specific thermodynamic state. We could envision continuing to take snapshots of our system to see other possible microstates. As you no doubt see, there would be such a staggeringly large number of microstates that taking individual snapshots of all of them is not feasible. Because we are examining such a large number of particles, 쑿 FIGURE 19.7 Ludwig Boltzmann’s gravestone. however, we can use the tools of statistics and probability to determine Boltzmann’s gravestone in Vienna is inscribed with his the total number of microstates for the thermodynamic state. (That is famous relationship between the entropy of a state, S, and where the statistical part of the name statistical thermodynamics comes the number of available microstates, W. (In Boltzmann’s in.) Each thermodynamic state has a characteristic number of mitime, “log” was used to represent the natural logarithm.) crostates associated with it, and we will use the symbol W for that number. Students sometimes have difficulty distinguishing between the state of a system and the microstates associated with the state. The difference is that state is used to describe the macroscopic view of our system as characterized, for example, by the pressure or temperature of a sample of gas. A microstate is a particular microscopic arrangement of the atoms or molecules of the system that corresponds to the given state of the system. Each of the snapshots we described is a microstate—the positions and kinetic energies of individual gas molecules will change from snapshot to snapshot, but each one is a possible arrangement of the collection of molecules corresponding to a single state. For macroscopically sized systems, such as a mole of gas, there is a very large number of microstates for each state—that is, W is generally an extremely large number. The connection between the number of microstates of a system, W, and the entropy of the system, S, is expressed in a beautifully simple equation developed by Boltzmann and engraved on his tombstone (씰 FIGURE 19.7): S = k ln W
[19.5] -23
In this equation, k is Boltzmann’s constant, 1.38 * 10 J>K. Thus, entropy is a measure of how many microstates are associated with a particular macroscopic state. GIVE IT SOME THOUGHT What is the entropy of a system that has only a single microstate?
From Equation 19.5, we see that the entropy change accompanying any process is ¢S = k ln Wfinal - k ln Winitial = k ln
Wfinal Winitial
[19.6]
Any change in the system that leads to an increase in the number of microstates 1Wfinal 7 Winitial2 leads to a positive value of ¢S: Entropy increases with the number of microstates of the system. Let’s consider two modifications to our ideal-gas sample and see how the entropy changes in each case. First, suppose we increase the volume of the system, which is analogous to allowing the gas to expand isothermally. A greater volume means a greater number of positions available to the gas atoms and therefore a greater number of microstates. The entropy therefore increases as the volume increases, as we saw in the “A Closer Look” box in Section 19.2.
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Second, suppose we keep the volume fixed but increase the temperature. How does this change affect the entropy of the system? Recall the distribution of molecular speeds presented in Figure 10.17(a). An increase in temperature increases the most probable speed of the molecules and also broadens the distribution of speeds. Hence, the molecules have a greater number of possible kinetic energies, and the number of microstates increases. Thus, the entropy of the system increases with increasing temperature.
Molecular Motions and Energy When a substance is heated, the motion of its molecules increases. In Section 10.7, we found that the average kinetic energy of the molecules of an ideal gas is directly proportional to the absolute temperature of the gas. That means the higher the temperature, the faster the molecules move and the more kinetic energy they possess. Moreover, hotter systems have a broader distribution of molecular speeds, as Figure 10.17(a) shows. The particles of an ideal gas are idealized points with no volume and no bonds, however, points that we visualize as flitting around through space. Any real molecule can undergo three kinds of more complex motion. The entire molecule can move in one direction, which is the simple motion we visualize for an ideal particle and see in a macroscopic object, such as a thrown baseball. We call such movement translational motion. The molecules in a gas have more freedom of translational motion than those in a liquid, which have more freedom of translational motion than the molecules of a solid. A real molecule can also undergo vibrational motion, in which the atoms in the molecule move periodically toward and away from one another, and rotational motion, in which the molecule spins about an axis. 쑼 FIGURE 19.8 shows the vibrational motions and one of the rotational motions possible for the water molecule. These different forms of motion are ways in which a molecule can store energy, and we refer to the various forms collectively as the motional energy of the molecule. GIVE IT SOME THOUGHT What kinds of motion can a molecule undergo that a single atom cannot?
The vibrational and rotational motions possible in real molecules lead to arrangements that a single atom can’t have. A collection of real molecules therefore has a greater number of possible microstates than does the same number of ideal-gas particles. In general, the number of microstates possible for a system increases with an increase in volume, an increase in temperature, or an increase in the number of molecules because any of these changes increases the possible positions and kinetic energies of the molecules making up the system. We will also see that the number of microstates increases as the complexity of the molecule increases because there are more vibrational motions available. Chemists have several ways of describing an increase in the number of microstates possible for a system and therefore an increase in the entropy for the system. Each way seeks to capture a sense of the increased freedom of motion that causes molecules to spread out when not restrained by physical barriers or chemical bonds.
GO FIGURE
Describe another possible rotational motion for this molecule.
씰 FIGURE 19.8 Vibrational and rotational motions in a water molecule.
Vibrations
Rotation
SECTION 19.3
Molecular Interpretation of Entropy
The most common way for describing an increase in entropy is as an increase in the randomness, or disorder, of the system. Another way likens an entropy increase to an increased dispersion (spreading out) of energy because there is an increase in the number of ways the positions and energies of the molecules can be distributed throughout the system. Each description (randomness or energy dispersal) is conceptually helpful if applied correctly.
Making Qualitative Predictions About ¢S It is usually not difficult to estimate qualitatively how the entropy of a system changes during a simple process. As noted earlier, an increase in either the temperature or the volume of a system leads to an increase in the number of microstates, and hence an increase in the entropy. One more factor that correlates with number of microstates is the number of independently moving particles. We can usually make qualitative predictions about entropy changes by focusing on these factors. For example, when water vaporizes, the molecules spread out into a larger volume. Because they occupy a larger volume, there is an increase in their freedom of motion, giving rise to a greater number of possible microstates, and hence an increase in entropy. Now consider the phases of water. In ice, hydrogen bonding leads to the rigid structure shown in 쑼 FIGURE 19.9. Each molecule in the ice is free to vibrate, but its translational and rotational motions are much more restricted than in liquid water. Although there are hydrogen bonds in liquid water, the molecules can more readily move about relative to one another (translation) and tumble around (rotation). During melting, therefore, the number of possible microstates increases and so does the entropy. In water vapor, the molecules are essentially independent of one another and have their full range of translational, vibrational, and rotational motions. Thus, water vapor has an even greater number of possible microstates and therefore a higher entropy than liquid water or ice.
GO FIGURE
In which phase are water molecules least able to have rotational motion? Increasing entropy
Ice
Rigid, crystalline structure Motion restricted to vibration only Smallest number of microstates
Liquid water
Water vapor
Increased freedom with respect to translation
Molecules spread out, essentially independent of one another
Free to vibrate and rotate
Complete freedom for translation, vibration, and rotation
Larger number of microstates
Largest number of microstates 쑿 FIGURE 19.9 Entropy and the phases of water. The larger the number of possible microstates, the higher the entropy of the system.
797
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ⴙ
ⴚ ⴚ ⴙ
ⴚ ⴙ ⴚ ⴙ ⴚ ⴙ ⴚ ⴙ ⴚ ⴙ ⴚ ⴚ ⴙ ⴚ ⴙ ⴚ ⴙ ⴚ ⴙ ⴚ ⴚ ⴙ ⴚ ⴙ ⴚ ⴙ ⴚ
쑿 FIGURE 19.10 Entropy changes when an ionic solid dissolves in water. The ions become more spread out and disordered, but the water molecules that hydrate the ions become less disordered.
When an ionic solid dissolves in water, a mixture of water and ions replaces the pure solid and pure water, as shown for KCl in 씱 FIGURE 19.10. The ions in the liquid move in a volume that is larger than the volume in which they were able to move in the crystal lattice and so possess more motional energy. This increased motion might lead us to conclude that the entropy of the system has increased. We have to be careful, however, because some of the water molecules have lost some freedom of motion because they are now held around the ions as water of hydration. •(Section 13.1) These water molecules are in a more ordered state than before because they are now confined to the immediate environment of the ions. Therefore, the dissolving of a salt involves both a disordering process (the ions become less confined) and an ordering process (some water molecules become more confined). The disordering processes are usually dominant, and so the overall effect is an increase in the randomness of the system when most salts dissolve in water. The same ideas apply to chemical reactions. Consider the reaction between nitric oxide gas and oxygen gas to form nitrogen dioxide gas: 2 NO(g2 + O2(g2 ¡ 2 NO2(g2
[19.7]
which results in a decrease in the number of molecules—three molecules of gaseous reactants form two molecules of gaseous products (씰 FIGURE 19.11). The formation of new N ¬ O bonds reduces the motions of the atoms in the system. The formation of new bonds decreases the number of degrees of freedom, or forms of motion, available to the atoms. That is, the atoms are less free to move in random fashion because of the formation of new bonds. The decrease in the number of molecules and the resultant decrease in motion result in fewer possible microstates and therefore a decrease in the entropy of the system. In summary, we generally expect the entropy of a system to increase for processes in which 1. Gases form from either solids or liquids. 2. Liquids or solutions form from solids. 3. The number of gas molecules increases during a chemical reaction. SAMPLE EXERCISE 19.3
Predicting the Sign of ¢S
Predict whether ¢S is positive or negative for each process, assuming each occurs at constant temperature: (a) H2O(l2 ¡ H2O(g2 (b) Ag +1aq2 + Cl-1aq2 ¡ AgCl1s2 (c) 4 Fe1s2 + 3 O21g2 ¡ 2 Fe 2O31s2 (d) N21g2 + O21g2 ¡ 2 NO1g2 SOLUTION Analyze We are given four reactions and asked to predict the sign of ¢S for each. Plan We expect ¢S to be positive if there is an increase in temperature, increase in volume, or increase in number of gas particles. The question states that the temperature is constant, and so we need to concern ourselves only with volume and number of particles. Solve (a) Evaporation involves a large increase in volume as liquid changes to gas. One mole of water (18 g) occupies about 18 mL as a liquid and if it could exist as a gas at STP it would occupy 22.4 L. Because the molecules are distributed throughout a much larger volume in the gaseous state, an increase in motional freedom accompanies vaporization and ¢S is positive. (b) In this process, ions, which are free to move throughout the volume of the solution, form a solid, in which they are confined to a smaller volume and restricted to more highly constrained positions. Thus, ¢S is negative. (c) The particles of a solid are confined to specific locations and have fewer ways to move (fewer microstates) than do the molecules of a gas. Because O2 gas is converted into part of the solid product Fe2O3, ¢S is negative.
SECTION 19.3
Molecular Interpretation of Entropy
(d) The number of moles of reactant gases is the same as the number of moles of product gases, and so the entropy change is expected to be small. The sign of ¢S is impossible to predict based on our discussions thus far, but we can predict that ¢S will be close to zero. PRACTICE EXERCISE Indicate whether each process produces an increase or decrease in the entropy of the system: (a) (b) (c) (d)
799
GO FIGURE
What major factor leads to a decrease in entropy as the reaction shown takes place?
CO21s2 ¡ CO21g2 CaO1s2 + CO21g2 ¡ CaCO3(s2 HCl1g2 + NH31g2 ¡ NH4Cl1s2 2 SO21g2 + O21g2 ¡ 2 SO31g2
Answers: (a) increase, (b) decrease, (c) decrease, (d) decrease
SAMPLE EXERCISE 19.4
Predicting Relative Entropies
In each pair, choose the system that has greater entropy and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl(g) at 25 °C, (b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25 °C, (c) 1 mol of HCl(g) or 1 mol of Ar(g) at 298 K. SOLUTION Analyze We need to select the system in each pair that has the greater entropy. Plan We examine the state of each system and the complexity of the molecules it contains. Solve (a) HCl(g) has the higher entropy because the particles in gases are more disordered and have more freedom of motion than the particles in solids. (b) When these two systems are at the same pressure, the sample containing 2 mol of HCl has twice the number of molecules as the sample containing 1 mol. Thus, the 2-mol sample has twice the number of microstates and twice the entropy. (c) The HCl system has the higher entropy because the number of ways in which an HCl molecule can store energy is greater than the number of ways in which an Ar atom can store energy. (Molecules can rotate and vibrate; atoms cannot.) PRACTICE EXERCISE Choose the system with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100 °C and 0.5 atm, (b) 1 mol of H2O(s) at 0 °C or 1 mol of H2O(l) at 25 °C, (c) 1 mol of H2(g) at STP or 1 mol of SO2(g) at STP, (d) 1 mol of N2O4(g) at STP or 2 mol of NO2(g) at STP. Answers: (a) 1 mol of H2(g) at 100 °C and 0.5 atm, (b) 1 mol of H2O(l) at 25 °C, (c) 1 mol of SO2(g) at STP, (d) 2 mol of NO2(g) at STP
The Third Law of Thermodynamics If we decrease the thermal energy of a system by lowering the temperature, the energy stored in translational, vibrational, and rotational motion decreases. As less energy is stored, the entropy of the system decreases. If we keep lowering the temperature, do we reach a state in which these motions are essentially shut down, a point described by a single microstate? This question is addressed by the third law of thermodynamics, which states that the entropy of a pure crystalline substance at absolute zero is zero: S10 K2 = 0. Consider a pure crystalline solid. At absolute zero, the individual atoms or molecules in the lattice would be perfectly ordered and as well defined in position as they could be. Because none of them would have thermal motion, there is only one possible microstate. As a result, Equation 19.5 becomes S = k ln W = k ln 1 = 0. As the temperature is increased from absolute zero, the atoms or molecules in the crystal gain energy in the form of vibrational motion about their lattice positions. This means that the degrees of freedom and the entropy both increase. What happens to the entropy, however, as we continue to heat the crystal? We consider this important question in the next section. GIVE IT SOME THOUGHT If you are told that the entropy of a system is zero, what do you know about the system?
2 NO(g) O2(g)
2 NO2(g)
쑿 FIGURE 19.11 Entropy decreases when NO(g) is oxidized by O2(g) to NO2(g). A decrease in the number of gaseous molecules leads to a decrease in the entropy of the system.
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CHEMISTRY AND LIFE ENTROPY AND HUMAN SOCIETY The laws of thermodynamics have profound implications for our existence. In the “Chemistry Put to Work” box on page 192, we examined some of the scientific and political challenges of using biofuels as a major energy source to maintain our lifestyles. That discussion builds around the first law of thermodynamics, namely, that energy is conserved. We therefore have important decisions to make as to energy production and consumption. The second law of thermodynamics is also relevant in discussions about our existence and about our ability and desire to advance as a civilization. Any living organism is a complex, highly organized, well-ordered system. Our entropy content is much lower than it would be if we were completely decomposed into carbon dioxide, water, and several other simple chemicals. Does this mean that our existence is a violation of the second law? No, because the thousands of chemical reactions necessary to produce and maintain human life have caused a very large increase in the entropy of the rest of the universe. Thus, as the second law requires, the overall entropy change during the lifetime of a human, or any other living system, is positive. In addition to being complex living systems ourselves, we humans are masters of producing order in the world around us. As shown in the chapter-opening photograph, we build impressive, highly ordered structures and buildings. We manipulate and order matter at the nanoscale level in order to produce the technological breakthroughs that have become so commonplace in the twenty-first century (씰 FIGURE 19.12). We use tremendous quantities of raw materials to produce highly ordered materials—iron, copper, and a host of other metals from their ores, silicon for computer chips from sand, polymers from fossil fuel feedstocks, and so forth. In so doing, we expend a great deal of energy to, in essence, “fight” the second law of thermodynamics. For every bit of order we produce, however, we produce an even greater amount of disorder. Petroleum, coal, and natural gas are burned to provide the energy necessary for us to achieve highly ordered structures, but their combustion increases the entropy of the universe by releasing CO2(g), H2O(g), and heat. Oxide and sulfide ores release CO2(g) and SO2(g) that spread throughout our
atmosphere. Thus, even as we strive to create more impressive discoveries and greater order in our society, we drive the entropy of the universe higher, as the second law says we must. We humans are, in effect, using up our storehouse of energyrich materials to create order and advance technology. As noted in Chapter 5, we must learn to harness new energy sources, such as solar energy, before we exhaust the supplies of readily available energy of other kinds.
쑿 FIGURE 19.12 Fighting the second law. Creating complex structures, such as the skyscrapers in the chapter-opening photograph, requires that we use energy to produce order while knowing that we are increasing the entropy of the universe. Modern cellular telephones, with their detailed displays and complex circuitry are an example on a smaller scale of the impressive order that human ingenuity achieves.
|
19.4 ENTROPY CHANGES IN CHEMICAL REACTIONS In Section 5.5 we discussed how calorimetry can be used to measure ¢H for chemical reactions. No comparable method exists for measuring ¢S for a reaction. However, because the third law establishes a zero point for entropy, we can use experimental measurements to determine the absolute value of the entropy, S. To see schematically how this is done, let’s review in greater detail the variation in the entropy of a substance with temperature. We know that the entropy of a pure crystalline solid at 0 K is zero and that the entropy increases as the temperature of the crystal is increased. 씰 FIGURE 19.13 shows that the entropy of the solid increases steadily with increasing temperature up to the melting point of the solid. When the solid melts, the atoms or molecules are free to move about the entire volume of the sample. The added degrees of freedom increase the randomness of the substance, thereby increasing its entropy. We therefore see a sharp increase in the entropy at the melting point. After all the solid has melted, the temperature again increases and with it, the entropy.
SECTION 19.4
Entropy Changes in Chemical Reactions
801
GO FIGURE
Why does the plot show vertical jumps at the melting and boiling points? Liquid
Entropy, S
Solid
Gas
Boiling
Melting 0
0 Temperature (K)
쑿 FIGURE 19.13 Entropy increases with increasing temperature.
At the boiling point of the liquid, another abrupt increase in entropy occurs. We can understand this increase as resulting from the increased volume available to the atoms or molecules as they enter the gaseous state. When the gas is heated further, the entropy increases steadily as more energy is stored in the translational motion of the gas atoms or molecules. Another change that occurs at higher temperatures is that the distribution of molecular speeds is skewed toward higher values. •[Figure 10.17(a)] The expansion of the range of speeds leads to increased kinetic energy and increased disorder and, hence, increased entropy. The conclusions we reach in examining Figure 19.13 are consistent with what we noted earlier: Entropy generally increases with increasing temperature because the increased motional energy leads to a greater number of possible microstates. Entropy plots such as Figure 19.13 can be obtained by carefully measuring how the heat capacity of a substance •(Section 5.5) varies with temperature, and we can use the data to obtain the absolute entropies at different temperatures. (The theory and methods used for these measurements and calculations are beyond the scope of this text.) Entropies are usually tabulated as molar quantities, in units of joules per mole-kelvin 1J>mol-K2. Molar entropies for substances in their standard states are known as standard molar entropies and denoted S°. The standard state for any substance is defined as the pure substance at 1 atm pressure.* 씰 TABLE 19.1 lists the values of S° for a number of substances at 298 K; Appendix C gives a more extensive list. We can make several observations about the S° values in Table 19.1: 1. Unlike enthalpies of formation, standard molar entropies of elements at the reference temperature of 298 K are not zero. 2. The standard molar entropies of gases are greater than those of liquids and solids, consistent with our interpretation of experimental observations, as represented in Figure 19.13. 3. Standard molar entropies generally increase with increasing molar mass. 4. Standard molar entropies generally increase with an increasing number of atoms in the formula of a substance. Point 4 is related to the molecular motion discussed in Section 19.3. In general, the number of degrees of freedom for a molecule increases with increasing number of *The standard pressure used in thermodynamics is no longer 1 atm but rather is based on the SI unit for pressure, the pascal (Pa). The standard pressure is 105 Pa, a quantity known as a bar: 1 bar = 105 Pa = 0.987 atm. Because 1 bar differs from 1 atm by only 1.3%, we will continue to refer to the standard pressure as 1 atm.
TABLE 19.1 • Standard Molar Entropies of Selected Substances at 298 K Substance
S° (J>mol-K)
H2(g) N2(g) O2(g) H2O(g) NH3(g) CH3OH(g) C6H6(g)
130.6 191.5 205.0 188.8 192.5 237.6 269.2
H2O(l) CH3OH(l) C6H6(l)
69.9 126.8 172.8
Li(s) Na(s) K(s) Fe(s) FeCl3(s) NaCl(s)
29.1 51.4 64.7 27.23 142.3 72.3
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GO FIGURE
What might you expect for the value of S° for butane, C4H10?
Methane, CH4 S 186.3 J/mol-K
Ethane, C2H6 S 229.6 J/mol-K
Propane, C3H8 S 270.3 J/mol-K
쑿 FIGURE 19.14 Entropy increases with increasing molecular complexity.
atoms, and thus the number of possible microstates also increases. 쑿 FIGURE 19.14 compares the standard molar entropies of three hydrocarbons in the gas phase. Notice how the entropy increases as the number of atoms in the molecule increases. The entropy change in a chemical reaction equals the sum of the entropies of the products minus the sum of the entropies of the reactants: ¢S° = g nS°1products2- g mS°1reactants2
[19.8]
As in Equation 5.31, the coefficients n and m are the coefficients in the balanced chemical equation for the reaction. SAMPLE EXERCISE 19.5
Calculating ¢S° from Tabulated Entropies
Calculate the change in the standard entropy of the system, ¢S°, for the synthesis of ammonia from N2(g) and H2(g) at 298 K: N21g2 + 3 H21g2 ¡ 2 NH31g2 SOLUTION Analyze We are asked to calculate the standard entropy change for the synthesis of NH3(g) from its constituent elements. Plan We can make this calculation using Equation 19.8 and the standard molar entropy values in Table 19.1 and Appendix C. Solve Using Equation 19.8, we have
¢S° = 2S°(NH32-3S°(N22 + 3S°(H224
Substituting the appropriate S° values from Table 19.1 yields
¢S° = 12 mol21192.5 J>mol-K2-311 mol21191.5 J>mol-K2 + 13 mol21130.6 J>mol-K24 = -198.3 J>K
Check: The value for ¢S° is negative, in agreement with our qualitative prediction based on the decrease in the number of molecules of gas during the reaction. PRACTICE EXERCISE Using the standard molar entropies in Appendix C, calculate the standard entropy change, ¢S°, for the following reaction at 298 K: Al 2O31s2 + 3 H21g2 ¡ 2 Al1s2 + 3 H2O1g2 Answer: 180.39 J>K
Entropy Changes in the Surroundings We can use tabulated absolute entropy values to calculate the standard entropy change in a system, such as a chemical reaction, as just described. But what about the entropy change in the surroundings? We encountered this situation in Section 19.2, but it is good to revisit it now that we are examining chemical reactions.
SECTION 19.5
Gibbs Free Energy
803
We should recognize that the surroundings for any system serve essentially as a large, constant-temperature heat source (or heat sink if the heat flows from the system to the surroundings). The change in entropy of the surroundings depends on how much heat is absorbed or given off by the system. For an isothermal process, the entropy change of the surroundings is given by -qsys ¢Ssurr = T Because in a constant-pressure process, qsys is simply the enthalpy change for the reaction, ¢H, we can write - ¢Hsys ¢Ssurr = [19.9] T For the reaction in Sample Exercise 19.5, qsys is the enthalpy change for the reaction under standard conditions, ¢H °, so the changes in entropy will be standard entropy changes, ¢S°. Therefore, using the procedures described in Section 5.7, we have ° = 2¢H f°3NH31g24 - 3¢H f°3H21g24 - ¢H f°3N21g24 ¢H rxn
= 21-46.19 kJ2 - 310 kJ2 - 10 kJ2 = -92.38 kJ The negative value tells us that at 298 K the formation of ammonia from H2(g) and N2(g) is exothermic. The surroundings absorb the heat given off by the system, which means an increase in the entropy of the surroundings: ° = ¢Ssurr
92.38 kJ = 0.310 kJ>K = 310 J>K 298 K
Notice that the magnitude of the entropy gained by the surroundings is greater than that lost by the system, calculated as -198.3 J>K in Sample Exercise 19.5. The overall entropy change for the reaction is ° = ¢Ssys ° + ¢Ssurr ° = -198.3 J>K + 310 J>K = 112 J>K ¢Suniv ° is positive for any spontaneous reaction, this calculation indicates that Because ¢Suniv when NH3(g), H2(g), and N2(g) are together at 298 K in their standard states (each at 1 atm pressure), the reaction moves spontaneously toward formation of NH3(g). Keep in mind that while the thermodynamic calculations indicate that formation of ammonia is spontaneous, they do not tell us anything about the rate at which ammonia is formed. Establishing equilibrium in this system within a reasonable period requires a catalyst, as discussed in Section 15.7. GIVE IT SOME THOUGHT If a process is exothermic, does the entropy of the surroundings (1) always increase, (2) always decrease, or (3) sometimes increase and sometimes decrease, depending on the process?
19.5 | GIBBS FREE ENERGY We have seen examples of endothermic processes that are spontaneous, such as the dissolution of ammonium nitrate in water. •(Section 13.1) We learned in our discussion of the solution process that a spontaneous process that is endothermic must be accompanied by an increase in the entropy of the system. However, we have also encountered processes that are spontaneous and yet proceed with a decrease in the entropy of the system, such as the highly exothermic formation of sodium chloride from its constituent elements. •(Section 8.2) Spontaneous processes that result in a decrease in the system’s entropy are always exothermic. Thus, the spontaneity of a reaction seems to involve two thermodynamic concepts, enthalpy and entropy. How can we use ¢H and ¢S to predict whether a given reaction occurring at constant temperature and pressure will be spontaneous? The means for doing so was first developed by the American mathematician J. Willard Gibbs (1839–1903). Gibbs (씰 FIGURE 19.15)
쑿 FIGURE 19.15 Josiah Willard Gibbs. Gibbs was the first person to be awarded a Ph.D. in science from an American university (Yale, 1863). From 1871 until his death, he held the chair of mathematical physics at Yale. He developed much of the theoretical foundation that led to the development of chemical thermodynamics.
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Chemical Thermodynamics
proposed a new state function, now called the Gibbs free energy (or just free energy), G, and defined as G = H - TS
[19.10]
where T is the absolute temperature. For an isothermal process, the change in the free energy of the system, ¢G, is ¢G = ¢H - T¢S
[19.11]
Under standard conditions, this equation becomes ¢G° = ¢H° - T¢S°
[19.12]
To see how the state function G relates to reaction spontaneity, recall that for a reaction occurring at constant temperature and pressure ¢Suniv = ¢Ssys + ¢Ssurr = ¢Ssys + a
- ¢Hsys T
b
where we have used Equation 19.9 to substitute for ¢Ssurr. Multiplying both sides by -T gives us -T¢Suniv = ¢Hsys - T¢Ssys
GO FIGURE
Are the processes that move a system toward equilibrium spontaneous or nonspontaneous?
Comparing Equations 19.11 and 19.13, we see that in a process occurring at constant temperature and pressure, the free-energy change, ¢G, is equal to -T¢Suniv. We know that for spontaneous processes, ¢Suniv is always positive and, therefore, -T¢Suniv is always negative. Thus, the sign of ¢G provides us with extremely valuable information about the spontaneity of processes that occur at constant temperature and pressure. If both T and P are constant, the relationship between the sign of ¢G and the spontaneity of a reaction is as follows:
Potential energy
1. If ¢G 6 0, the reaction is spontaneous in the forward direction. 2. If ¢G = 0, the reaction is at equilibrium. 3. If ¢G 7 0, the reaction in the forward direction is nonspontaneous (work must be done to make it occur) but the reverse reaction is spontaneous. Equilibrium position in valley
Position
Reactants Free energy
[19.13]
Equilibrium mixture Products
Course of reaction 쑿 FIGURE 19.16 Potential energy and free energy. An analogy is shown between the gravitational potential-energy change of a boulder rolling down a hill and the freeenergy change in a spontaneous reaction.
It is more convenient to use ¢G as a criterion for spontaneity than to use ¢Suniv because ¢G relates to the system alone and avoids the complication of having to examine the surroundings. An analogy is often drawn between the free-energy change during a spontaneous reaction and the potential-energy change when a boulder rolls down a hill (씱 FIGURE 19.16). Potential energy in a gravitational field “drives” the boulder until it reaches a state of minimum potential energy in the valley. Similarly, the free energy of a chemical system decreases until it reaches a minimum value. When this minimum is reached, a state of equilibrium exists. In any spontaneous process carried out at constant temperature and pressure, the free energy always decreases. To illustrate these ideas, let’s return to the Haber process for the synthesis of ammonia from nitrogen and hydrogen, which we discussed extensively in Chapter 15: N21g2 + 3 H21g2 Δ 2 NH31g2 Imagine that we have a reaction vessel that allows us to maintain a constant temperature and pressure and that we have a catalyst that allows the reaction to proceed at a reasonable rate. What happens when we charge the vessel with a certain number of moles of N2 and three times that number of moles of H2? As we saw in Figure 15.3, the N2 and H2 react spontaneously to form NH3 until equilibrium is achieved. Similarly, Figure 15.3 shows that if we charge the vessel with pure NH3, it decomposes spontaneously to N2 and H2 until equilibrium is reached. In each case the free energy of the system gets progressively lower and lower as the reaction moves toward equilibrium, which represents a minimum in the free energy. We illustrate these cases in 씰 FIGURE 19.17.
SECTION 19.5
Gibbs Free Energy
805
GO FIGURE
Why are the spontaneous processes shown sometimes said to be “downhill” in free energy? N2(g) 3 H2(g)
2 NH3(g)
Free energy
Spontaneous Spontaneous
QK
Pure N 2 H2
QK
Equilibrium mixture (Q K, G 0)
Pure NH3
GIVE IT SOME THOUGHT What are the criteria for spontaneity a. in terms of entropy and b. in terms of free energy?
This is a good time to remind ourselves of the significance of the reaction quotient, Q, for a system that is not at equilibrium. •(Section 15.6) Recall that when Q 6 K, there is an excess of reactants relative to products and the reaction proceeds spontaneously in the forward direction to reach equilibrium, as noted in Figure 19.17. When Q 7 K, the reaction proceeds spontaneously in the reverse direction. At equilibrium Q = K. SAMPLE EXERCISE 19.6
Calculating Free-Energy Change from ¢H°, T, and ¢S°
Calculate the standard free-energy change for the formation of NO(g) from N2(g) and O2(g) at 298 K: N21g2 + O21g2 ¡ 2 NO1g2 given that ¢H° = 180.7 kJ and ¢S° = 24.7 J>K. Is the reaction spontaneous under these conditions? SOLUTION Analyze We are asked to calculate ¢G° for the indicated reaction (given ¢H°, ¢S°, and T) and to predict whether the reaction is spontaneous under standard conditions at 298 K. Plan To calculate ¢G°, we use Equation 19.12, ¢G° = ¢H° - T¢S°. To determine whether the reaction is spontaneous under standard conditions, we look at the sign of ¢G°.
씱 FIGURE 19.17 Free energy and approaching equilibrium. In the reaction N2 (g) + 3 H2 (g) Δ 2 NH3 (g), if the reaction mixture has too much N2 and H2 relative to NH3 (left), the equilibrium lies too far to the left (Q 6 K) and NH3 forms spontaneously. If there is too much NH3 in the mixture (right), the equilibrium lies too far to the right (Q 7 K) and the NH3 decomposes spontaneously into N2 and H2.
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Solve ¢G° = ¢H° - T¢S° = 180.7 kJ - 1298 K2124.7 J>K2a
1 kJ 103 J
b
= 180.7 kJ - 7.4 kJ = 173.3 kJ Because ¢G° is positive, the reaction is not spontaneous under standard conditions at 298 K. Comment Notice that we had to convert the units of the T¢S° term to kJ so that they could be added to the ¢H° term, whose units are kJ. PRACTICE EXERCISE Calculate ¢G° for a reaction for which ¢H° = 24.6 kJ and ¢S° = 132 J>K at 298 K. Is the reaction spontaneous under these conditions? Answer: ¢G° = -14.7 kJ; the reaction is spontaneous.
Standard Free Energy of Formation TABLE 19.2 • Conventions Used in Establishing Standard Free Energies State of Matter
Standard State
Solid Liquid Gas Solution Element
Pure solid Pure liquid 1 atm pressure 1 M concentration ¢G f° = 0 for element in standard state
Recall that we defined standard enthalpies of formation, ¢H f°, as the enthalpy change when a substance is formed from its elements under defined standard conditions. •(Section 5.7) We can define standard free energies of formation, ¢G f°, in a similar way. As is summarized in 씱 TABLE 19.2, standard state means 1 atm pressure for gases, the pure solid for solids, and the pure liquid for liquids. For substances in solution, the standard state is normally a concentration of 1 M. (In very accurate work it may be necessary to make certain corrections, but we need not worry about these.) The temperature usually chosen for purposes of tabulating data is 25 °C, but we will calculate ¢G° at other temperatures as well. Just as for the standard heats of formation, the free energies of elements in their standard states are set to zero. This arbitrary choice of a reference point has no effect on the quantity in which we are interested, which is the difference in free energy between reactants and products. A listing of standard free energies of formation is given in Appendix C. GIVE IT SOME THOUGHT What does the superscript ° indicate when associated with a thermodynamic quantity, as in ¢H°,¢S°, or ¢G°?
Standard free energies of formation are useful in calculating the standard freeenergy change for chemical processes. The procedure is analogous to the calculation of ¢H° (Equation 5.31) and ¢S° (Equation 19.8): ¢G° = a n¢G f°1products2 - a m¢G f°1reactants2
SAMPLE EXERCISE 19.7
[19.14]
Calculating Standard Free-Energy Change from Free Energies of Formation
(a) Use data from Appendix C to calculate the standard free-energy change for the reaction P4(g) + 6 Cl 2 (g) ¡ 4 PCl 3 (g) run at 298 K. (b) What is ¢G° for the reverse of this reaction? SOLUTION Analyze We are asked to calculate the free-energy change for a reaction and then to determine the free-energy change for the reverse reaction. Plan We look up the free-energy values for the products and reactants and use Equation 19.14: We multiply the molar quantities by the coefficients in the balanced equation and subtract the total for the reactants from that for the products.
SECTION 19.5
Solve (a) Cl2(g) is in its standard state, so ¢G f° is zero for this reactant. P4(g), however, is not in its standard state, so ¢G f° is not zero for this reactant. From the balanced equation and values from Appendix C, we have ° = 4 ¢G f° 3PCl 3(g)4 - ¢G f° 3P4(g)4 - 6 ¢G f° 3Cl 2(g)4 ¢G rxn = (4 mol)(-269.6 kJ>mol) - (1 mol)(24.4 kJ>mol) - 0 = - 1102.8 kJ
That ¢G° is negative tells us that a mixture of P4(g), Cl2(g), and PCl3(g) at 25 °C, each present at a partial pressure of 1 atm, would react spontaneously in the forward direction to form more PCl3. Remember, however, that the value of ¢G° tells us nothing about the rate at which the reaction occurs. (b) When we reverse the reaction, we reverse the roles of the reactants and products. Thus, reversing the reaction changes the sign of ¢G in Equation 19.14, just as reversing the reaction changes the sign of ¢H. •(Section 5.4) Hence, using the result from part (a), we have 4 PCl 3(g) ¡ P4(g) + 6 Cl 2(g)
¢G° = +1102.8 kJ
PRACTICE EXERCISE Use data from Appendix C to calculate ¢G° at 298 K for the combustion of methane: CH4(g) + 2 O2(g) ¡ CO2(g) + 2 H2O(g). Answer: -800.7 kJ SAMPLE EXERCISE 19.8
Estimating and Calculating ¢G°
In Section 5.7 we used Hess’s law to calculate ¢H° for the combustion of propane gas at 298 K: C3H8(g) + 5 O2(g) ¡ 3 CO2(g) + 4 H2O(l)
¢H° = -2220 kJ
(a) Without using data from Appendix C, predict whether ¢G° for this reaction is more negative or less negative than ¢H°. (b) Use data from Appendix C to calculate ¢G° for the reaction at 298 K. Is your prediction from part (a) correct? SOLUTION Analyze In part (a) we must predict the value for ¢G° relative to that for ¢H° on the basis of the balanced equation for the reaction. In part (b) we must calculate the value for ¢G° and compare this value with our qualitative prediction. Plan The free-energy change incorporates both the change in enthalpy and the change in entropy for the reaction (Equation 19.11), so under standard conditions ¢G° = ¢H° - T¢S° To determine whether ¢G° is more negative or less negative than ¢H°, we need to determine the sign of the term T¢S°. Because T is the absolute temperature, 298 K, it is always a positive number. We can predict the sign of ¢S° by looking at the reaction. Solve (a) The reactants are six molecules of gas, and the products are three molecules of gas and four molecules of liquid. Thus, the number of molecules of gas has decreased significantly during the reaction. By using the general rules discussed in Section 19.3, we expect a decrease in the number of gas molecules to lead to a decrease in the entropy of the system—the products have fewer possible microstates than the reactants. We therefore expect ¢S° and T¢S° to be negative. Because we are subtracting T¢S°, which is a negative number, we predict that ¢G° is less negative than ¢H°. (b) Using Equation 19.14 and values from Appendix C, we have ¢G° = 3¢G f° [CO2(g)] + 4¢G f° [H2O(l)] - ¢G f° [C3H8(g)] - 5¢G f° [O2(g)] = 3 mol( -394.4 kJ>mol) + 4 mol( -237.13 kJ>mol)1 mol( -23.47 kJ>mol) - 5 mol(0 kJ>mol) = -2108 kJ Notice that we have been careful to use the value of ¢G f° for H2O(l). As in calculating ¢H values, the phases of the reactants and products are important. As we predicted, ¢G° is less negative than ¢H° because of the decrease in entropy during the reaction. PRACTICE EXERCISE For the combustion of propane at 298 K, C3H8(g) + 5 O2(g) ¡ 3 CO2(g) + 4 H2O(g), do you expect ¢G° to be more negative or less negative than ¢H°? Answer: more negative
Gibbs Free Energy
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A CLOSER LOOK WHAT’S “FREE” ABOUT FREE ENERGY? The Gibbs free energy is a remarkable thermodynamic quantity. Because so many chemical reactions are carried out under conditions of near-constant pressure and temperature, chemists, biochemists, and engineers use the sign and magnitude of ¢G as exceptionally useful tools in the design of chemical and biochemical reactions. We will see examples of the usefulness of ¢G throughout the remainder of this chapter and this text. Two common questions often arise when one first learns about the Gibbs free energy: Why does the sign of ¢G tell us about the spontaneity of reactions? And what is “free” about free energy? We address these two questions here by using concepts discussed in Chapter 5 and earlier in this chapter. In Section 19.2 we saw that the second law of thermodynamics governs the spontaneity of processes. In order to apply the second law (Equations 19.4), however, we must determine ¢Suniv, which is often difficult to evaluate. When T and P are constant, however, we can relate ¢Suniv to the changes in entropy and enthalpy of just the system by substituting the Equation 19.9 expression for ¢Ssurr in Equation 19.4: ¢Suniv = ¢Ssys + ¢Ssurr = ¢Ssys + a
- ¢Hsys T
b
[19.15]
(constant T, P) Thus, at constant temperature and pressure, the second law becomes Reversible process: Irreversible process:
¢Suniv = ¢Ssys ¢Suniv = ¢Ssys -
¢Hsys T ¢Hsys T
= 0 [19.16] 7 0 (constant T, P)
Now we can see the relationship between ¢Gsys (which we call simply ¢G) and the second law. From Equation 19.11 we know that ¢G = ¢Hsys - T¢Ssys. If we multiply Equations 19.16 by T and rearrange, we reach the following conclusion: Reversible process: ¢G = ¢Hsys - T ¢Ssys = 0 Irrversible process: ¢G = ¢Hsys - T ¢Ssys 6 0
[19.17]
Equations 19.17 allow us to use the sign of ¢G to conclude whether a reaction is spontaneous, nonspontaneous, or at equilibrium. When ¢G 6 0, a process is irreversible and, therefore, spontaneous. When ¢G = 0, the process is reversible and, therefore, at equilibrium. If a process has ¢G 7 0, then the reverse process will have ¢G 6 0; thus, the process as written is nonspontaneous but its reverse reaction will be irreversible and spontaneous. The magnitude of ¢G is also significant. A reaction for which ¢G is large and negative, such as the burning of gasoline, is much more capable of doing work on the surroundings than is a reaction for which ¢G is small and negative, such as ice melting at room temperature. In fact, thermodynamics tells us that the change in free energy for a process, ¢G, equals the maximum useful work that can be done by the system on its surroundings in a spontaneous process occurring at constant temperature and pressure: ¢G = - wmax
[19.18]
(Remember our sign convention from Table 5.1: Work done by a system is negative.) In other words, ¢G gives the theoretical limit to how much work can be done by a process. The relationship in Equation 19.18 explains why ¢G is called free energy—it is the portion of the energy change of a spontaneous reaction that is free to do useful work. The remainder of the energy enters the environment as heat. For example, the theoretical maximum work obtained for the combustion of gasoline is given by the value of ¢G for the combustion reaction. On average, standard internal combustion engines are inefficient in utilizing this potential work—more than 60% of the potential work is lost (primarily as heat) in converting the chemical energy of the gasoline to mechanical energy to move the vehicle (쑼 FIGURE 19.18). When other losses are considered—idling time, braking, aerodynamic drag, and so forth—only about 15% of the potential work from the gasoline is used to move the car. Advances in automobile design—such as hybrid technology, efficient diesel engines, and new lightweight materials—have the potential to increase the percentage of useful work obtained from the gasoline. For nonspontaneous processes 1¢G 7 02, the free-energy change is a measure of the minimum amount of work that must be done to cause the process to occur. In actuality, we always need to do more than this theoretical minimum amount because of the inefficiencies in the way the changes occur.
(constant T, P)
Useful work (15%)
씰 FIGURE 19.18 Energy losses in automobiles. Very little of the chemical energy of gasoline is actually used as work to move a typical automobile.
Chemical energy of gasoline, G
Wasted energy 85% (heat loss from engine, braking, air resistance, and so forth)
SECTION 19.6
Free Energy and Temperature
19.6 | FREE ENERGY AND TEMPERATURE Tabulations of ¢G f°, such as those in Appendix C, make it possible to calculate ¢G° for reactions at the standard temperature of 25 °C, but we are often interested in examining reactions at other temperatures. To see how ¢G is affected by temperature, let’s look again at Equation 19.11: ¢G = ¢H - T¢S = ¢H + (-T¢S) Enthalpy term
Entropy term
Notice that we have written the expression for ¢G as a sum of two contributions, an enthalpy term, ¢H, and an entropy term, -T¢S. Because the value of -T¢S depends directly on the absolute temperature T, ¢G varies with temperature. We know that the enthalpy term, ¢H, can be either positive or negative and that T is positive at all temperatures other than absolute zero. The entropy term, -T¢S, can also be positive or negative. When ¢S is positive, which means the final state has greater randomness (a greater number of microstates) than the initial state, the term -T¢S is negative. When ¢S is negative, -T¢S is positive. The sign of ¢G, which tells us whether a process is spontaneous, depends on the signs and magnitudes of ¢H and -T¢S. The various combinations of ¢H and -T¢S signs are given in 쑼 TABLE 19.3. Note in Table 19.3 that when ¢H and -T¢S have opposite signs, the sign of ¢G depends on the magnitudes of these two terms. In these instances temperature is an important consideration. Generally, ¢H and ¢S change very little with temperature. However, the value of T directly affects the magnitude of -T¢S. As the temperature increases, the magnitude of -T¢S increases, and this term becomes relatively more important in determining the sign and magnitude of ¢G. As an example, let’s consider once more the melting of ice to liquid water at 1 atm: H2O(s) ¡ H2O(l)
¢H 7 0, ¢S 7 0
This process is endothermic, which means that ¢H is positive. Because the entropy increases during the process, ¢S is positive, which makes -T¢S negative. At temperatures below 0 °C (273 K), the magnitude of ¢H is greater than that of -T¢S. Hence, the positive enthalpy term dominates, and ¢G is positive. This positive value of ¢G means that ice melting is not spontaneous at T 6 0 °C, just as our everyday experience tells us; rather, the reverse process, the freezing of liquid water into ice, is spontaneous at these temperatures. What happens at temperatures greater than 0 °C? As T increases, so does the magnitude of -T¢S. When T 7 0 °C, the magnitude of -T¢S is greater than the magnitude of ¢H, which means that the -T¢S term dominates and ¢G is negative. The negative value of ¢G tells us that ice melting is spontaneous at T 7 0 °C. At the normal melting point of water, T = 0 °C, the two phases are in equilibrium. Recall that ¢G = 0 at equilibrium; at T = 0 °C, ¢H and -T¢S are equal in magnitude and opposite in sign, so they cancel and give ¢G = 0. TABLE 19.3 • How Signs of ¢H and ¢S Affect Reaction Spontaneity ¢H
¢S
-T¢S
¢G = ¢H - T¢S
Reaction Characteristics
Example
+ -
+ -
+ +
+ + or -
Spontaneous at all temperatures Nonspontaneous at all temperatures Spontaneous at low T; nonspontaneous at high T
2 O31g2 ¡ 3 O21g2 3 O21g2 ¡ 2 O31g2 H2O1l2 ¡ H2O1s2
+
+
-
+ or -
Spontaneous at high T; nonspontaneous at low T
H2O1s2 ¡ H2O1l2
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CHAPTER 19
Chemical Thermodynamics GIVE IT SOME THOUGHT The normal boiling point of benzene is 80 °C. At 100 °C and 1 atm, which term is greater in magnitude for the vaporization of benzene, ¢H or T¢S?
Our discussion of the temperature dependence of ¢G is also relevant to standard free-energy changes. We can calculate the values of ¢H° and ¢S° at 298 K from the data in Appendix C. If we assume that these values do not change with temperature, we can then use Equation 19.12 to estimate ¢G° at temperatures other than 298 K. SAMPLE EXERCISE 19.9
Determining the Effect of Temperature on Spontaneity
The Haber process for the production of ammonia involves the equilibrium N2(g) + 3 H2(g) Δ 2 NH3(g) Assume that ¢H° and ¢S° for this reaction do not change with temperature. (a) Predict the direction in which ¢G° for the reaction changes with increasing temperature. (b) Calculate ¢G° at 25 °C and 500 °C. SOLUTION Analyze In part (a) we are asked to predict the direction in which ¢G° changes as temperature increases. In part (b) we need to determine ¢G° for the reaction at two temperatures. Plan We can answer part (a) by determining the sign of ¢S for the reaction and then using that information to analyze Equation 19.12. In part (b) we first calculate ¢H° and ¢S° for the reaction using data in Appendix C and then use Equation 19.12 to calculate ¢G°. Solve (a) The temperature dependence of ¢G° comes from the entropy term in Equation 19.12, ¢G° = ¢H° - T¢S°. We expect ¢S° for this reaction to be negative because the number of molecules of gas is smaller in the products. Because ¢S° is negative, -T¢S° is positive and increases with increasing temperature. As a result, ¢G° becomes less negative (or more positive) with increasing temperature. Thus, the driving force for the production of NH3 becomes smaller with increasing temperature. (b) We calculated ¢H° for this reaction in Sample Exercise 15.14 and ¢S° in Sample Exercise 19.5: ¢H° = -92.38 kJ and ¢S° = -198.3 J>K. If we assume that these values do not change with temperature, we can calculate ¢G° at any temperature by using Equation 19.12. At T = 25 °C = 298 K, we have ¢G° = -92.38 kJ - (298 K)(-198.3 J>K)a
1 kJ b 1000 J
= -92.38 kJ + 59.1 kJ = -33.3 kJ At T = 500 °C = 773 K, we have ¢G ° = -92.38 kJ - (773 K)a -198.3
1 kJ J ba b K 1000 J
= -92.38 kJ + 153 kJ = 61 kJ Notice that we had to convert the units of -T¢S° to kJ in both calculations so that this term can be added to the ¢H° term, which has units of kJ. Comment Increasing the temperature from 298 K to 773 K changes ¢G° from -33.3 kJ to +61 kJ. Of course, the result at 773 K assumes that ¢H° and ¢S° do not change with temperature. Although these values do change slightly with temperature, the result at 773 K should be a reasonable approximation. The positive increase in ¢G° with increasing T agrees with our prediction in part (a). Our result indicates that in a mixture of N2(g), H2(g), and NH3(g), each present at a partial pressure of 1 atm, the N2(g) and H2(g) react spontaneously at 298 K to form more NH3(g). At 773 K, the positive value of ¢G° tells us that the reverse reaction is spontaneous. Thus, when the mixture of these gases, each at a partial pressure of 1 atm, is heated to 773 K, some of the NH3(g) spontaneously decomposes into N2(g) and H2(g). PRACTICE EXERCISE (a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate ¢H° and ¢S° at 298 K for the reaction 2 SO2(g) + O2(g) ¡ 2 SO3(g). (b) Use your values from part (a) to estimate ¢G° at 400 K. Answers: (a) ¢H° = -196.6 kJ, ¢S° = -189.6 J>K; (b) ¢G° = -120.8 kJ
SECTION 19.7
Free Energy and the Equilibrium Constant
811
|
19.7 FREE ENERGY AND THE EQUILIBRIUM CONSTANT In Section 19.5 we saw a special relationship between ¢G and equilibrium: For a system at equilibrium, ¢G = 0. We have also seen how to use tabulated thermodynamic data to calculate values of the standard free-energy change, ¢G°. In this final section, we learn two more ways in which we can use free energy to analyze chemical reactions: using ¢G° to calculate ¢G under nonstandard conditions and relating the values of ¢G° and K for a reaction.
Free Energy Under Nonstandard Conditions The set of standard conditions for which ¢G° values pertain is given in Table 19.2. Most chemical reactions occur under nonstandard conditions. For any chemical process, the relationship between the free-energy change under standard conditions, ¢G°, and the free-energy change under any other conditions, ¢G, is given by ¢G = ¢G° + RT ln Q
[19.19]
In this equation R is the ideal-gas constant, 8.314 J>mol-K; T is the absolute temperature; and Q is the reaction quotient for the reaction mixture of interest. •(Section 15.6) Under standard conditions, the concentrations of all the reactants and products are equal to 1. Thus, under standard conditions Q = 1, ln Q = 0, and Equation 19.19 reduces to ¢G = ¢G° under standard conditions, as it should.
SAMPLE EXERCISE 19.10
Relating ¢G to a Phase Change at Equilibrium
(a) Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCl4(l). (b) What is the value of ¢G° for the equilibrium in part (a)? (c) Use data from Appendix C and Equation 19.12 to estimate the normal boiling point of CCl4. SOLUTION Analyze (a) We must write a chemical equation that describes the physical equilibrium between liquid and gaseous CCl4 at the normal boiling point. (b) We must determine the value of ¢G° for CCl4, in equilibrium with its vapor at the normal boiling point. (c) We must estimate the normal boiling point of CCl4, based on available thermodynamic data. Solve (a) The normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm:
Plan (a) The chemical equation is the change of state from liquid to gas. For (b), we need to analyze Equation 19.19 at equilibrium (¢G = 0), and for (c) we can use Equation 19.12 to calculate T when ¢G = 0.
CCl 4(l) Δ CCl 4(g) P = 1 atm
(b) At equilibrium, ¢G = 0. In any normal boiling-point equilibrium, both liquid and vapor are in their standard state of pure liquid and vapor at 1 atm (Table 19.2). Consequently, Q = 1, ln Q = 0, and ¢G = ¢G ° for this process. We conclude that ¢G° = 0 for the equilibrium representing the normal boiling point of any liquid. (We would also find that ¢G° = 0 for the equilibria relevant to normal melting points and normal sublimation points.) (c) Combining Equation 19.12 with the result from part (b), we see that the equality at the normal boiling point, Tb, of ¢G° = ¢H° - Tb ¢S° = 0 CCl4(l) (or any other pure liquid) is Solving the equation for Tb, we obtain
Tb = ¢H° > ¢S°
Strictly speaking, we need the values of ¢H ° and ¢S° for the CCl4(l)–CCl4(g) equilibrium at the normal boiling point to do this calculation. However, we can estimate the boiling point by using the values of ¢H ° and ¢S° for CCl4 at 298 K, which we obtain from Appendix C and Equations 5.31 and 19.8:
¢H° = (1 mol)(- 106.7 kJ>mol) - (1 mol)(-139.3 kJ>mol) = +32.6 kJ ¢S° = (1 mol)(309.4 J>mol-K) - (1 mol)(214.4 J>mol-K) = +95.0 J>K
As expected, the process is endothermic (¢H 7 0) and produces a gas, thus increasing the entropy (¢S 7 0). We now use these values to estimate Tb for CCl4(l):
Tb =
32.6 kJ 1000 J ¢H° = a ba b = 343 K = 70 °C ¢S° 95.0 J>K 1 kJ
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Note that we have used the conversion factor between joules and kilojoules to make the units of ¢H° and ¢S° match. Check The experimental normal boiling point of CCl4(l) is 76.5 °C. The small deviation of our estimate from the experimental value is due to the assumption that ¢H° and ¢S° do not change with temperature. PRACTICE EXERCISE Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br2(l). (The experimental value is given in Figure 11.5.) Answer: 330 K
When the concentrations of reactants and products are nonstandard, we must calculate Q in order to determine ¢G. We illustrate how this is done in Sample Exercise 19.11. At this stage in our discussion, therefore, it becomes important to note the units used to calculate Q when using Equation 19.19. The convention used for standard states is used when applying this equation: In determining the value of Q, the concentrations of gases are always expressed as partial pressures in atmospheres and solutes are expressed as their concentrations in molarities. SAMPLE EXERCISE 19.11
Calculating the Free-Energy Change under Nonstandard Conditions
Calculate ¢G at 298 K for a mixture of 1.0 atm N2, 3.0 atm H2, and 0.50 atm NH3 being used in the Haber process: N2(g) + 3 H2(g) Δ 2 NH3(g) SOLUTION Analyze We are asked to calculate ¢G under nonstandard conditions. Plan We can use Equation 19.19 to calculate ¢G. Doing so requires that we calculate the value of the reaction quotient Q for the specified partial pressures, for which we use the partial-pressures form of Equation 15.23: Q = 3D4d3E4e>3A4a3B4b. We then use a table of standard free energies of formation to evaluate ¢G°. Solve The partial-pressures form of Equation 15.23 gives PNH3 2 (0.50)2 = 9.3 * 10- 3 Q = 3 = PN2 PH2 (1.0)(3.0)3 In Sample Exercise 19.9 we calculated ¢G° = -33.3 kJ for this reaction. We will have to change the units of this quantity in applying Equation 19.19, however. For the units in Equation 19.19 to work out, we will use kJ>mol as our units for ¢G°, where “per mole” means “per mole of the reaction as written.” Thus, ¢G° = -33.3 kJ>mol implies per 1 mol of N2, per 3 mol of H2, and per 2 mol of NH3. We now use Equation 19.19 to calculate ¢G for these nonstandard conditions: ¢G = ¢G° + RT ln Q = 1-33.3 kJ>mol2 + 18.314 J>mol-K21298 K211 kJ>1000 J2 ln19.3 * 10-32 = 1 -33.3 kJ>mol2 + 1-11.6 kJ>mol2 = -44.9 kJ>mol Comment We see that ¢G becomes more negative as the pressures of N2, H2, and NH3 are changed from 1.0 atm (standard conditions, ¢G°) to 1.0 atm, 3.0 atm, and 0.50 atm, respectively. The larger negative value for ¢G indicates a larger “driving force” to produce NH3. We would make the same prediction based on Le Châtelier’s principle. • (Section 15.7) Relative to standard conditions, we have increased the pressure of a reactant (H2) and decreased the pressure of the product (NH3). Le Châtelier’s principle predicts that both changes shift the reaction to the product side, thereby forming more NH3. PRACTICE EXERCISE Calculate ¢G at 298 K for the Haber reaction if the reaction mixture consists of 0.50 atm N2, 0.75 atm H2, and 2.0 atm NH3. Answer: -26.0 kJ>mol
SECTION 19.7
Free Energy and the Equilibrium Constant
813
Relationship Between ¢G° and K We can now use Equation 19.19 to derive the relationship between ¢G° and the equilibrium constant, K. At equilibrium, ¢G = 0 and Q = K. Thus, at equilibrium, Equation 19.19 transforms as follows: ¢G = ¢G° + RT ln Q 0 = ¢G° + RT ln K ¢G° = -RT ln K
[19.20]
By solving Equation 19.20 for K, we obtain an expression that allows us to calculate K if we know the value of ¢G°: ln K =
¢G° -RT
K = e
-¢G°>RT
[19.21]
As usual, we must be careful in our choice of units. In Equations 19.20 and 19.21 we again express ¢G° in kJ>mol. In the equilibrium-constant expression, we use atmospheres for gas pressures, molarities for solutions, and solids, liquids, and solvents do not appear in the expression. •(Section 15.4) Thus, the equilibrium constant is Kp for gas-phase reactions and Kc for reactions in solution. •(Section 15.2) From Equation 19.20 we see that if ¢G° is negative, ln K must be positive, which means K 7 1. Therefore, the more negative ¢G° is, the larger K is. Conversely, if ¢G° is positive, ln K is negative, which means K 6 1. 씰 TABLE 19.4 summarizes these relationships.
SAMPLE EXERCISE 19.12
TABLE 19.4 • Relationship between ¢G ° and K at 298 K ¢G ° (kJ/ mol)
K
+200 +100 +50 +10 +1.0 0 -1.0 -10 -50 -100 -200
8.7 3.0 1.7 1.8 6.7 1.0 1.5 5.7 5.8 3.4 1.1
* * * * *
10-36 10-18 10-9 10-2 10-1
* * * *
101 108 1017 1035
Calculating an Equilibrium Constant from ¢G°
The standard free-energy change for the Haber process at 25 °C was obtained in Sample Exercise 19.9 for the Haber reaction: N21g2 + 3 H21g2 Δ 2 NH31g2
¢G° = -33.3 kJ>mol = -33,300 J>mol
Use this value of ¢G° to calculate the equilibrium constant for the process at 25 °C. SOLUTION Analyze We are asked to calculate K for a reaction, given ¢G°. Plan We can use Equation 19.21 to calculate K. Solve Remembering to use the absolute temperature for T in Equation 19.21 and the form of R that matches our units, we have Comment This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mixture at 25 °C. The equilibrium constants for the Haber reaction at temperatures in the range 300 °C to 600 °C, given in Table 15.2, are much smaller than the value at 25 °C. Clearly, a low-temperature equilibrium favors the production of ammonia more than a hightemperature one. Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature.
K = e -¢G°>RT = e -1-33,300 J>mol2>(8.314 J>mol-K21298 K2 = e13.4 = 7 * 105 Remember Thermodynamics can tell us the direction and extent of a reaction but tells us nothing about the rate at which it will occur. If a catalyst were found that would permit the reaction to proceed at a rapid rate at room temperature, high pressures would not be needed to force the equilibrium toward NH3.
PRACTICE EXERCISE Use data from Appendix C to calculate ¢G° and K at 298 K for the reaction H21g2 + Br21l2 Δ 2 HBr1g2. Answer: ¢G° = - 106.4 kJ>mol, K = 4 * 1018
814
CHAPTER 19
Chemical Thermodynamics
CHEMISTRY AND LIFE DRIVING NONSPONTANEOUS REACTIONS Many desirable chemical reactions, including a large number that are central to living systems, are nonspontaneous as written. For example, consider the extraction of copper metal from the mineral chalcocite, which contains Cu2S. The decomposition of Cu2S to its elements is nonspontaneous: Cu 2S1s2 ¡ 2 Cu1s2 + S1s2
C6H12O61s2 + 6 O21g2 ¡ 6 CO21g2 + 6 H2O1l2
¢G° = +86.2 kJ
¢G° = -2880 kJ
Because ¢G° is very positive, we cannot obtain Cu(s) directly via this reaction. Instead, we must find some way to “do work” on the reaction to force it to occur as we wish. We can do this by coupling the reaction to another one so that the overall reaction is spontaneous. For example, we can envision the S(s) reacting with O2(g) to form SO2(g): S1s2 + O21g2 ¡ SO21g2
biochemical reactions that are essential for the formation and maintenance of highly ordered biological structures are not spontaneous. These necessary reactions are made to occur by coupling them with spontaneous reactions that release energy. The metabolism of food is the usual source of the free energy needed to do the work of maintaining biological systems. For example, complete oxidation of the sugar glucose, C6H12O6, to CO2 and H2O yields substantial free energy:
¢G° = -300.4 kJ
By coupling these reactions, we can extract much of the copper metal via a spontaneous reaction: Cu 2S1s2 + O21g2 ¡ 2 Cu1s2 + SO21g2 ¢G° = 1+86.2 kJ2 + 1-300.4 kJ2 = -214.2 kJ In essence, we have used the spontaneous reaction of S(s) with O2(g) to provide the free energy needed to extract the copper metal from the mineral. Biological systems employ the same principle of using spontaneous reactions to drive nonspontaneous ones. Many of the
This energy can be used to drive nonspontaneous reactions in the body. However, a means is necessary to transport the energy released by glucose metabolism to the reactions that require energy. One way, shown in 쑼 FIGURE 19.19, involves the interconversion of adenosine triphosphate (ATP) and adenosine diphosphate (ADP), molecules that are related to the building blocks of nucleic acids. The conversion of ATP to ADP releases free energy 1¢G° = -30.5 kJ2 that can be used to drive other reactions. In the human body the metabolism of glucose occurs via a complex series of reactions, most of which release free energy. The free energy released during these steps is used in part to reconvert lowerenergy ADP back to higher-energy ATP. Thus, the ATP–ADP interconversions are used to store energy during metabolism and to release it as needed to drive nonspontaneous reactions in the body. If you take a course in biochemistry, you will have the opportunity to learn more about the remarkable sequence of reactions used to transport free energy throughout the human body. RELATED EXERCISES: 19.102 and 19.103
Glucose (C6H12O6)
Cell constituents Cell
Glucose oxidation
High free energy
Cellular development
ATP 씰 FIGURE 19.19 Schematic representation of free-energy changes during cell metabolism. The oxidation of glucose to CO2 and H2O produces free energy that is then used to convert ADP into the more energetic ATP. The ATP is then used, as needed, as an energy source to drive nonspontaneous reactions, such as the conversion of simple molecules into more complex cell constituents.
Free energy released by oxidation of glucose converts ADP to ATP
CO2 ⴙ H2O
ADP Low free energy
Free energy released by ATP converts simple molecules to more complex ones
Simpler molecules
SECTION 19.7
SAMPLE INTEGRATIVE EXERCISE
Free Energy and the Equilibrium Constant
Putting Concepts Together
Consider the simple salts NaCl(s) and AgCl(s). We will examine the equilibria in which these salts dissolve in water to form aqueous solutions of ions: NaCl1s2 Δ Na + 1aq2 + Cl-1aq2 AgCl1s2 Δ Ag + 1aq2 + Cl-1aq2 (a) Calculate the value of ¢G° at 298 K for each of the preceding reactions. (b) The two values from part (a) are very different. Is this difference primarily due to the enthalpy term or the entropy term of the standard free-energy change? (c) Use the values of ¢G° to calculate the Ksp values for the two salts at 298 K. (d) Sodium chloride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part (c)? (e) How will ¢G° for the solution process of these salts change with increasing T? What effect should this change have on the solubility of the salts? SOLUTION (a) We will use Equation 19.14 along with ¢G f° values from Appendix C to calculate the ° values for each equilibrium. (As we did in Section 13.1, we use the subscript “soln” to ¢G soln indicate that these are thermodynamic quantities for the formation of a solution.) We find ° 1NaCl2 = 1-261.9 kJ>mol2 + 1-131.2 kJ>mol2 - 1-384.0 kJ>mol2 ¢G soln
= -9.1 kJ>mol ° 1AgCl2 = 1+77.11 kJ>mol2 + 1-131.2 kJ>mol2 - 1-109.70 kJ>mol2 ¢G soln
= +55.6 kJ>mol ° as the sum of an enthalpy term, ¢H soln ° , and an entropy term, (b) We can write ¢G soln ° : ¢G soln ° = ¢H soln ° + 1-T¢Ssoln ° 2. We can calculate the values of ¢H soln ° and ¢Ssoln ° -T¢Ssoln ° at T = 298 K. All these by using Equations 5.31 and 19.8. We can then calculate -T¢Ssoln calculations are now familiar to us. The results are summarized in the following table:
Salt
° ¢Hsoln
° ¢Ssoln
° T¢Ssoln
NaCl AgCl
+3.6 kJ>mol +65.7 kJ>mol
+43.2 kJ>mol-K +34.3 kJ>mol-K
-12.9 kJ>mol -10.2 kJ>mol
The entropy terms for the solution of the two salts are very similar. That seems sensible because each solution process should lead to a similar increase in randomness as the salt dissolves, forming hydrated ions. •(Section 13.1) In contrast, we see a very large difference ° is in the enthalpy term for the solution of the two salts. The difference in the values of ¢G soln ° . dominated by the difference in the values of ¢H soln (c) The solubility product, Ksp, is the equilibrium constant for the solution process. ° by using Equation 19.21: •(Section 17.4) As such, we can relate Ksp directly to ¢G soln Ksp = e -¢G°soln>RT We can calculate the Ksp values in the same way we applied Equation 19.21 in Sample Exercise ° values we obtained in part (a), remembering to convert them from 19.12. We use the ¢G soln kJ>mol to J>mol: NaCl: Ksp = 3Na + 1aq243Cl-1aq24 = e -1-91002>318.3142129824 = e + 3.7 = 40 AgCl: Ksp = 3Ag + 1aq243Cl-1aq24 = e -1 + 55,6002>318.3142129824 = e -22.4 = 1.9 * 10-10 The value calculated for the Ksp of AgCl is very close to that listed in Appendix D. (d) A soluble salt is one that dissolves appreciably in water. • (Section 4.2) The Ksp value for NaCl is greater than 1, indicating that NaCl dissolves to a great extent. The Ksp value for AgCl is very small, indicating that very little dissolves in water. Silver chloride should indeed be considered an insoluble salt. (e) As we expect, the solution process has a positive value of ¢S for both salts (see the table in ° , is negative. If we aspart b). As such, the entropy term of the free-energy change, -T¢Ssoln ° and ¢Ssoln ° do not change much with temperature, then an increase in T will sume that ¢H soln ° more negative. Thus, the driving force for dissolution of the salts will inserve to make ¢G soln crease with increasing T, and we therefore expect the solubility of the salts to increase with increasing T. In Figure 13.18 we see that the solubility of NaCl (and that of nearly any other salt) increases with increasing temperature. • (Section 13.3)
815
816
CHAPTER 19
Chemical Thermodynamics
CHAPTER SUMMARY AND KEY TERMS Most reactions and chemical processes have an inherent directionality: They are spontaneous in one direction and nonspontaneous in the reverse direction. The spontaneity of a process is related to the thermodynamic path the system takes from the initial state to the final state. In a reversible process, both the system and its surroundings can be restored to their original state by exactly reversing the change. In an irreversible process the system cannot return to its original state without there being a permanent change in the surroundings. Any spontaneous process is irreversible. A process that occurs at a constant temperature is said to be isothermal. SECTION 19.1
SECTION 19.2 The spontaneous nature of processes is related to a thermodynamic state function called entropy, denoted S. For a process that occurs at constant temperature, the entropy change of the system is given by the heat absorbed by the system along a reversible path, divided by the temperature: ¢S = qrev>T . The way entropy controls the spontaneity of processes is given by the second law of thermodynamics, which governs the change in the entropy of the universe, ¢Suniv = ¢Ssys + ¢Ssurr. The second law states that in a reversible process ¢Suniv = 0; in an irreversible (spontaneous) process ¢Suniv 7 0. Entropy values are usually expressed in units of joules per kelvin, J>K. SECTION 19.3 A particular combination of motions and locations of the atoms and molecules of a system at a particular instant is called a microstate. The entropy of a system is a measure of its randomness or disorder. The entropy is related to the number of microstates, W, corresponding to the state of the system: S = k ln W. Molecules can undergo three kinds of motion: In translational motion the entire molecule moves in space. Molecules can also undergo vibrational motion, in which the atoms of the molecule move toward and away from one another in periodic fashion, and rotational motion, in which the entire molecule spins like a top. The number of available microstates, and therefore the entropy, increases with an increase in volume, temperature, or motion of molecules because any of these changes increases the possible motions and locations of the molecules. As a result, entropy generally increases when liquids or solutions are formed from solids, gases are formed from either solids or liquids, or the number of molecules of gas increases during a chemical reaction. The third law of thermodynamics states that the entropy of a pure crystalline solid at 0 K is zero.
The third law allows us to assign entropy values for substances at different temperatures. Under standard conditions the entropy of a mole of a substance is called its standard molar entropy, denoted S°. From tabulated values of S° , we can calculate the entropy change for any process under standard conditions. For an isothermal process, the entropy change in the surroundings is equal to - ¢H>T.
SECTION 19.4
SECTION 19.5 The Gibbs free energy (or just free energy), G, is a thermodynamic state function that combines the two state functions enthalpy and entropy: G = H - TS. For processes that occur at constant temperature, ¢G = ¢H - T¢S. For a process occurring at constant temperature and pressure, the sign of ¢G relates to the spontaneity of the process. When ¢G is negative, the process is spontaneous. When ¢G is positive, the process is nonspontaneous but the reverse process is spontaneous. At equilibrium the process is reversible and ¢G is zero. The free energy is also a measure of the maximum useful work that can be performed by a system in a spontaneous process. The standard free-energy change, ¢G°, for any process can be calculated from tabulations of standard free energies of formation, ¢G f°, which are defined in a fashion analogous to standard enthalpies of formation, ¢G f°. The value of ¢G f° for a pure element in its standard state is defined to be zero.
The values of ¢H and ¢S generally do not vary much with temperature. Therefore, the dependence of ¢G with temperature is governed mainly by the value of T in the expression ¢G = ¢H - T¢S. The entropy term -T¢S has the greater effect on the temperature dependence of ¢G and, hence, on the spontaneity of the process. For example, a process for which ¢H 7 0 and ¢S 7 0, such as the melting of ice, can be nonspontaneous 1¢G 7 02 at low temperatures and spontaneous 1¢G 6 02 at higher temperatures. Under nonstandard conditions ¢G is related to ¢G° and the value of the reaction quotient, Q: ¢G = ¢G° + RT ln Q. At equilibrium 1¢G = 0 , Q = K2, ¢G° = -RT ln K. Thus, the standard free-energy change is directly related to the equilibrium constant for the reaction. This relationship expresses the temperature dependence of equilibrium constants. SECTIONS 19.6 AND 19.7
KEY SKILLS • Understand the meaning of spontaneous process, reversible process, irreversible process, and isothermal process. (Section 19.1) • State the second law of thermodynamics. (Section 19.2) • Explain how the entropy of a system is related to the number of possible microstates. (Section 19.3) • Describe the kinds of molecular motion that a molecule can possess. (Section 19.3) • Predict the sign of ¢S for physical and chemical processes. (Section 19.3) • State the third law of thermodynamics. (Section 19.3) • Calculate standard entropy changes for a system from standard molar entropies. (Section 19.4) • Calculate entropy changes in the surroundings for isothermal processes. (Section 19.4) • Calculate the Gibbs free energy from the enthalpy change and entropy change at a given temperature. (Section 19.5) • Use free-energy changes to predict whether reactions are spontaneous. (Section 19.5) • Calculate standard free-energy changes using standard free energies of formation. (Section 19.5) • Predict the effect of temperature on spontaneity given ¢H and ¢S. (Section 19.6) • Calculate ¢G under nonstandard conditions. (Section 19.7) • Relate ¢G° and equilibrium constant. (Section 19.7)
Exercises
817
KEY EQUATIONS • ¢S =
qrev T
1constant T2
[19.2]
Relating entropy change to the heat absorbed or released in a reversible process
[19.4]
The second law of thermodynamics
• S = k ln W
[19.5]
Relating entropy to the number of microstates
• ¢S° = a nS°1products2 - a mS°1reactants2
[19.8]
Calculating the standard entropy change from standard molar entropies
[19.9]
The entropy change of the surroundings for a process at constant temperature and pressure
• ¢G = ¢H - T¢S
[19.11]
Calculating the Gibbs free-energy change from enthalpy and entropy changes at constant temperature
• ¢G° = a n¢G f°1products2 - a m¢G f°1reactants2
[19.14]
Calculating the standard free-energy change from standard free energies of formation
• Reversible process: Irreversible process:
• ¢Ssurr =
¢Suniv = ¢Ssys + ¢Ssurr = 0 ¢Suniv = ¢Ssys + ¢Ssurr 7 0
f
- ¢Hsys T
• Reversible process: Irreversible process:
¢G = ¢Hsys - T¢Ssys = 0 ¢G = ¢Hsys - T¢Ssys 6 0
f
[19.17]
Relating the free-energy change to the reversibility of a process at constant temperature and pressure
• ¢G = -wmax
[19.18]
Relating the free-energy change to the maximum work a process can perform
• ¢G = ¢G° + RT ln Q
[19.19]
Calculating free-energy change under nonstandard conditions
• ¢G° = -RT ln K
[19.20]
Relating the standard free-energy change and the equilibrium constant
EXERCISES VISUALIZING CONCEPTS 19.1 Two different gases occupy the two bulbs shown here. Consider the process that occurs when the stopcock is opened, assuming the gases behave ideally. (a) Draw the final (equilibrium) state. (b) Predict the signs of ¢H and ¢S for the process. (c) Is the process that occurs when the stopcock is opened a reversible one? (d) How does the process affect the entropy of the surroundings? [Sections 19.1 and 19.2]
Vaporized C2H4F2 19.2 As shown here, one type of computer keyboard cleaner contains liquefied 1,1-difluoroethane (C2H4F2), which is a gas at atmospheric pressure. When the nozzle is squeezed, the 1,1-difluoroethane vaporizes out of the nozzle at high pressure, blowing dust out of objects. (a) Based on your experience, is the vaporization a spontaneous process at room temperature? (b) Defining the 1,1-difluoroethane as the system, do you expect qsys for the process to be positive or negative? Explain. (c) Predict whether ¢S is positive or negative for this process. (d) Given your answers to (a), (b), and (c), do you think the operation of this product depends more on heat flow or more on entropy change?
Liquefied C2H4F2
818
CHAPTER 19
Chemical Thermodynamics
19.3 (a) What are the signs of ¢S and ¢H for the process depicted here? (b) How might temperature affect the sign of ¢G? (c) If energy can flow in and out of the system to maintain a constant temperature during the process, what can you say about the entropy change of the surroundings as a result of this process? [Sections 19.2 and 19.5]
19.7 The accompanying diagram shows how ¢H (red line) and T¢S (blue line) change with temperature for a hypothetical reaction. (a) What is the significance of the point at 300 K, where ¢H and T¢S are equal? (b) In what temperature range is this reaction spontaneous? [Section 19.6]
H
TS 19.4 Predict the sign of ¢S accompanying this reaction. Explain your choice. [Section 19.3]
300 K Temperature 19.8 The accompanying diagram shows how ¢G for a hypothetical reaction changes as temperature changes. (a) At what temperature is the system at equilibrium? (b) In what temperature range is the reaction spontaneous? (c) Is ¢H positive or negative? (d) Is ¢S positive or negative? [Sections 19.5 and 19.6]
19.5 The accompanying diagram shows how entropy varies with temperature for a substance that is a gas at the highest temperature shown. (a) What processes correspond to the entropy increases along the vertical lines labeled 1 and 2? (b) Why is the entropy change for 2 larger than that for 1? [Section 19.3]
250 K Temperature
2 Entropy, S
G 0
19.9 Consider a reaction A21g2 + B21g2 Δ 2 AB1g2, with atoms of A shown in red in the diagram and atoms of B shown in blue. (a) If Kc = 1, which box represents the system at equilibrium? (b) What is the sign of ¢G for any process in which the contents of a reaction vessel move to equilibrium? (c) Rank the boxes in order of increasing magnitude of ¢G for the reaction. [Sections 19.5 and 19.7]
1
Temperature (K) 19.6 Isomers are molecules that have the same chemical formula but different arrangements of atoms, as shown here for two isomers of pentane, C5H12. (a) Do you expect a significant difference in the enthalpy of combustion of the two isomers? Explain. (b) Which isomer do you expect to have the higher standard molar entropy? Explain. [Section 19.4]
CH2
CH2
CH2
CH3
CH3
C
(2)
x
CH3
CH3
n-pentane
neopentane
(3)
19.10 The accompanying diagram shows how the free energy, G, changes during a hypothetical reaction A1g2 + B1g2 ¡ C1g2. On the left are pure reactants, each at 1 atm, and on the right is the pure product, also at 1 atm. (a) What is the significance of the minimum in the plot? (b) What does the quantity x, shown on the right side of the diagram, represent? [Section 19.7]
G
CH3 CH3
(1)
Progress of reaction
Exercises
819
SPONTANEOUS PROCESSES (section 19.1) 19.11 Which of the following processes are spontaneous and which are nonspontaneous: (a) the ripening of a banana, (b) dissolution of sugar in a cup of hot coffee, (c) the reaction of nitrogen atoms to form N2 molecules at 25 °C and 1 atm, (d) lightning, (e) formation of CH4 and O2 molecules from CO2 and H2O at room temperature and 1 atm of pressure? 19.12 Which of the following processes are spontaneous: (a) the melting of ice cubes at -10 °C and 1 atm pressure; (b) separating a mixture of N2 and O2 into two separate samples, one that is pure N2 and one that is pure O2; (c) alignment of iron filings in a magnetic field; (d) the reaction of hydrogen gas with oxygen gas to form water vapor; (e) the dissolution of HCl(g) in water to form concentrated hydrochloric acid? 19.13 (a) Give two examples of endothermic processes that are spontaneous. (b) Give an example of a process that is spontaneous at one temperature but nonspontaneous at a different temperature. 19.14 The crystalline hydrate Cd1NO322 # 4 H2O1s2 loses water when placed in a large, closed, dry vessel: Cd1NO322 # 4 H2O1s2 ¡ Cd1NO3221s2 + 4 H2O1g2
This process is spontaneous and ¢H is positive. Is this process an exception to Bertholet’s generalization that all spontaneous changes are exothermic? Explain. 19.15 Consider the vaporization of liquid water to steam at a pressure of 1 atm. (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium? 19.16 The normal freezing point of n-octane (C8H18) is -57 °C. (a) Is the freezing of n-octane an endothermic or exothermic process? (b) In what temperature range is the freezing of n-octane a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any temperature at which liquid n-octane and solid n-octane are in equilibrium? Explain. 19.17 (a) What is special about a reversible process? (b) Suppose a reversible process is reversed, restoring the system to its original
state. What can be said about the surroundings after the process is reversed? (c) Under what circumstances will the vaporization of water to steam be a reversible process? (d) Are any of the processes that occur in the world around us reversible in nature? Explain. 19.18 (a) What is meant by calling a process irreversible? (b) After a particular irreversible process, the system is restored to its original state. What can be said about the condition of the surroundings after the system is restored to its original state? (c) Under what conditions will the condensation of a liquid be an irreversible process? 19.19 Consider a process in which an ideal gas changes from state 1 to state 2 in such a way that its temperature changes from 300 K to 200 K. (a) Describe how this change might be carried out while keeping the volume of the gas constant. (b) Describe how it might be carried out while keeping the pressure of the gas constant. (c) Does the change in ¢E depend on the particular pathway taken to carry out this change of state? Explain. 19.20 A system goes from state 1 to state 2 and back to state 1. (a) What is the relationship between the value of ¢E for going from state 1 to state 2 to that for going from state 2 back to state 1? (b) Without further information, can you conclude anything about the amount of heat transferred to the system as it goes from state 1 to state 2 as compared to that upon going from state 2 back to state 1? (c) Suppose the changes in state are reversible processes. Can you conclude anything about the work done by the system upon going from state 1 to state 2 as compared to that upon going from state 2 back to state 1? 19.21 Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is ¢E zero for the process? Explain. 19.22 Consider what happens when a sample of the explosive TNT (Section 8.8: “Chemistry Put to Work: Explosives and Alfred Nobel”) is detonated under atmospheric pressure. (a) Is the detonation a spontaneous process? (b) What is the sign of q for this process? (c) Can you determine whether w is positive, negative, or zero for the process? Explain. (d) Can you determine the sign of ¢E for the process? Explain.
ENTROPY AND THE SECOND LAW OF THERMODYNAMICS (section 19.2) 19.23 (a) How can we calculate ¢S for an isothermal process? (b) Does ¢S for a process depend on the path taken from the initial state to the final state of the system? Explain.
gallium solidifies to Ga(s) at its normal melting point, is ¢S positive or negative? (b) Calculate the value of ¢S when 60.0 g of Ga(l) solidifies at 29.8 °C.
19.24 Suppose we vaporize a mole of liquid water at 25 °C and another mole of water at 100 °C. (a) Assuming that the enthalpy of vaporization of water does not change much between 25 °C and 100 °C, which process involves the larger change in entropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? Explain.
19.27 (a) Express the second law of thermodynamics in words. (b) If the entropy of the system increases during a reversible process, what can you say about the entropy change of the surroundings? (c) In a certain spontaneous process the system undergoes an entropy change, ¢S = 42 J>K. What can you conclude about ¢Ssurr?
19.25 The normal boiling point of Br2(l) is 58.8 °C, and its molar enthalpy of vaporization is ¢Hvap = 29.6 kJ>mol. (a) When Br2(l) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of ¢S when 1.00 mol of Br2(l) is vaporized at 58.8 °C.
19.28 (a) Express the second law of thermodynamics as a mathematical equation. (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of ¢Ssurr? (c) During a certain reversible process, the surroundings undergo an entropy change, ¢Ssurr = -78 J>K. What is the entropy change of the system for this process?
19.26 The element gallium (Ga) freezes at 29.8 °C, and its molar enthalpy of fusion is ¢Hfus = 5.59 kJ>mol. (a) When molten
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19.29 (a) What sign for ¢S do you expect when the volume of 0.200 mol of an ideal gas at 27 °C is increased isothermally from an initial volume of 10.0 L? (b) If the final volume is 18.5 L, calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change? Explain.
19.30 (a) What sign for ¢S do you expect when the pressure on 0.600 mol of an ideal gas at 350 K is increased isothermally from an initial pressure of 0.750 atm? (b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change? Explain.
THE MOLECULAR INTERPRETATION OF ENTROPY (section 19.3) 19.31 For the isothermal expansion of a gas into a vacuum, ¢E = 0, q = 0, and w = 0. (a) Is this a spontaneous process? (b) Explain why no work is done by the system during this process. (c) In thermodynamics, what is the “driving force” for the expansion of the gas? 19.32 (a) What is the difference between a state and a microstate of a system? (b) As a system goes from state A to state B, its entropy decreases. What can you say about the number of microstates corresponding to each state? (c) In a particular spontaneous process, the number of microstates available to the system decreases. What can you conclude about the sign of ¢Ssurr? 19.33 How would each of the following changes affect the number of microstates available to a system: (a) increase in temperature, (b) decrease in volume, (c) change of state from liquid to gas? 19.34 (a) Using the heat of vaporization in Appendix B, calculate the entropy change for the vaporization of water at 25 °C and at 100 °C. (b) From your knowledge of microstates and the structure of liquid water, explain the difference in these two values. 19.35 (a) What do you expect for the sign of ¢S in a chemical reaction in which two moles of gaseous reactants are converted to three moles of gaseous products? (b) For which of the processes in Exercise 19.11 does the entropy of the system increase? 19.36 (a) In a chemical reaction two gases combine to form a solid. What do you expect for the sign of ¢S? (b) How does the entropy of the system change in the processes described in Exercise 19.12? 19.37 How does the entropy of the system change when (a) a solid melts, (b) a gas liquefies, (c) a solid sublimes? 19.38 How does the entropy of the system change when (a) the temperature of the system increases, (b) the volume of a gas increases, (c) equal volumes of ethanol and water are mixed to form a solution? 19.39 (a) State the third law of thermodynamics. (b) Distinguish between translational motion, vibrational motion, and
rotational motion of a molecule. (c) Illustrate these three kinds of motion with sketches for the HCl molecule. 19.40 (a) If you are told that the entropy of a certain system is zero, what do you know about the system and the temperature? (b) The energy of a gas is increased by heating it. Using CO2 as an example, illustrate the different ways in which additional energy can be distributed among the molecules of the gas. (c) CO2(g) and Ar(g) have nearly the same molar mass. At a given temperature, will they have the same number of microstates? Explain. 19.41 For each of the following pairs, choose the substance with the higher entropy per mole at a given temperature: (a) Ar(l) or Ar(g), (b) He(g) at 3 atm pressure or He(g) at 1.5 atm pressure, (c) 1 mol of Ne(g) in 15.0 L or 1 mol of Ne(g) in 1.50 L, (d) CO2(g) or CO2(s). 19.42 For each of the following pairs, indicate which substance possesses the larger standard entropy: (a) 1 mol of P4(g) at 300 °C, 0.01 atm, or 1 mol of As4(g) at 300 °C, 0.01 atm; (b) 1 mol of H2O(g) at 100 °C, 1 atm, or 1 mol of H2O(l) at 100 °C, 1 atm; (c) 0.5 mol of N2(g) at 298 K, 20-L volume, or 0.5 mol CH4(g) at 298 K, 20-L volume; (d) 100 g Na2SO4(s) at 30 °C or 100 g Na2SO4(aq) at 30 °C. 19.43 Predict the sign of the entropy change of the system for each of the following reactions: (a) N21g2 + 3 H21g2 ¡ 2 NH31g2 (b) CaCO31s2 ¡ CaO1s2 + CO21g2 (c) 3 C2H21g2 ¡ C6H61g2 (d) Al 2O31s2 + 3 H21g2 ¡ 2 Al1s2 + 3 H2O1g2 19.44 Predict the sign of ¢Ssys for each of the following processes: (a) Molten gold solidifies. (b) Gaseous Cl2 dissociates in the stratosphere to form gaseous Cl atoms. (c) Gaseous CO reacts with gaseous H2 to form liquid methanol, CH3OH. (d) Calcium phosphate precipitates upon mixing Ca(NO3)2(aq) and (NH4)3PO4(aq).
ENTROPY CHANGES IN CHEMICAL REACTIONS (section 19.4) 19.45 (a) Using Figure 19.13 as a model, sketch how the entropy of water changes as it is heated from -50 °C to 110 °C at sea level. Show the temperatures at which there are vertical increases in entropy. (b) Which process has the larger entropy change: melting ice or boiling water? Explain. 19.46 Propanol (C3H7OH) melts at -126.5 °C and boils at 97.4 °C. Draw a qualitative sketch of how the entropy changes as propanol vapor at 150 °C and 1 atm is cooled to solid propanol at - 150 °C and 1 atm. 19.47 In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) C2H2(g) or C2H6(g), (b) CO2(g) or CO(g)? Explain.
19.48 Cyclopropane and propylene are isomers (see Exercise 19.6) that both have the formula C3H6. Based on the molecular structures shown, which of these isomers would you expect to have the higher standard molar entropy at 25 °C? H
H
H C
H
H
H
C H
H
C
C C
H Cyclopropane
H
C
H H
Propylene
Exercises 19.49 Use Appendix C to compare the standard entropies at 25 °C for the following pairs of substances: (a) Sc(s) and Sc(g), (b) NH3(g) and NH3(aq), (c) 1 mol P4(g) and 2 mol P2(g), (d) C(graphite) and C(diamond). In each case explain the difference in the entropy values. 19.50 Using Appendix C, compare the standard entropies at 25 °C for the following pairs of substances: (a) CuO(s) and Cu2O(s), (b) 1 mol N2O4(g) and 2 mol NO2(g), (c) SiO2(s) and CO2(g), (d) CO(g) and CO2(g). For each pair, explain the difference in the entropy values. [19.51] The standard entropies at 298 K for certain of the group 4A elements are as follows: C(s, diamond) 2.43 J>mol-K, Si(s) 18.81 J>mol-K, Ge(s) 31.09 J>mol-K, and Sn(s) 51.818 J>mol-K. All but Sn have the diamond structure. How do you account for the trend in the S° values? [19.52] Three of the forms of elemental carbon are graphite, diamond, and buckminsterfullerene. The entropies at 298 K for graphite and diamond are listed in Appendix C. (a) Account
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for the difference in the S° values of graphite and diamond in light of their structures (Figure 12.30). (b) What would you expect for the S° value of buckminsterfullerene (Figure 12.47) relative to the values for graphite and diamond? Explain. 19.53 Using S° values from Appendix C, calculate ¢S° values for the following reactions. In each case account for the sign of ¢S°. (a) C2H41g2 + H21g2 ¡ C2H61g2 (b) N2O41g2 ¡ 2 NO21g2 (c) Be1OH22 1s2 ¡ BeO1s2 + H2O1g2 (d) 2 CH3OH1g2 + 3 O21g2 ¡ 2 CO21g2 + 4 H2O1g2 19.54 Calculate ¢S° values for the following reactions by using tabulated S° values from Appendix C. In each case explain the sign of ¢S°. (a) HNO31g2 + NH31g2 ¡ NH4NO31s2 (b) 2 Fe 2O31s2 ¡ 4 Fe1s2 + 3 O21g2 (c) CaCO31s, calcite2 + 2HCl1g2 ¡ CaCl 21s2 + CO21g2 + H2O1l2 (d) 3 C2H61g2 ¡ C6H61l2 + 6 H21g2
GIBBS FREE ENERGY (sections 19.5 and 19.6) 19.55 (a) For a process that occurs at constant temperature, express the change in Gibbs free energy in terms of changes in the enthalpy and entropy of the system. (b) For a certain process that occurs at constant T and P, the value of ¢G is positive. What can you conclude? (c) What is the relationship between ¢G for a process and the rate at which it occurs? 19.56 (a) What is the meaning of the standard free-energy change, ¢G°, as compared with ¢G? (b) For any process that occurs at constant temperature and pressure, what is the significance of ¢G = 0? (c) For a certain process, ¢G is large and negative. Does this mean that the process necessarily occurs rapidly? 19.57 For a certain chemical reaction, ¢H° = -35.4 kJ and ¢S° = -85.5 J>K. (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate ¢G° for the reaction at 298 K. (d) Is the reaction spontaneous at 298 K under standard conditions? 19.58 A certain reaction has ¢H° = +23.7 kJ and ¢S° = +52.4 J>K. (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate ¢G° for the reaction at 298 K. (d) Is the reaction spontaneous at 298 K under standard conditions? 19.59 Using data in Appendix C, calculate ¢H°, ¢S°, and ¢G° at 298 K for each of the following reactions. In each case show that ¢G° = ¢H° - T¢S°. (a) H21g2 + F21g2 ¡ 2 HF1g2 (b) C1s, graphite2 + 2 Cl 21g2 ¡ CCl 41g2 (c) 2 PCl 31g2 + O21g2 ¡ 2 POCl 31g2 (d) 2 CH3OH1g2 + H21g2 ¡ C2H61g2 + 2 H2O1g2 19.60 Use data in Appendix C to calculate ¢H°, ¢S°, and ¢G° at 25 °C for each of the following reactions. In each case show that ¢G° = ¢H° - T¢S°. (a) 2 Cr1s2 + 3 O21g2 ¡ 2 CrO31s2 (b) BaCO31s2 ¡ BaO1s2 + CO21g2 (c) 2 P1s2 + 10 HF1g2 ¡ 2 PF51g2 + 5 H21g2 (d) K1s2 + O21g2 ¡ KO21s2
19.61 Using data from Appendix C, calculate ¢G° for the following reactions. Indicate whether each reaction is spontaneous at 298 K under standard conditions. (a) 2 SO21g2 + O21g2 ¡ 2 SO31g2 (b) NO21g2 + N2O1g2 ¡ 3 NO1g2 (c) 6 Cl 21g2 + 2 Fe 2O31s2 ¡ 4 FeCl 31s2 + 3 O21g2 (d) SO21g2 + 2 H21g2 ¡ S1s2 + 2 H2O1g2 19.62 Using data from Appendix C, calculate the change in Gibbs free energy for each of the following reactions. In each case indicate whether the reaction is spontaneous at 298 K under standard conditions. (a) 2 Ag1s2 + Cl 21g2 ¡ 2 AgCl1s2 (b) P4O61s2 + 12 H21g2 ¡ 4 PH31g2 + 6 H2O1g2 (c) CH41g2 + 4 F21g2 ¡ CF41g2 + 4 HF1g2 (d) 2 H2O21l2 ¡ 2 H2O1l2 + O21g2 19.63 Octane (C8H18) is a liquid hydrocarbon at room temperature that is the primary constituent of gasoline. (a) Write a balanced equation for the combustion of C8H18(l) to form CO2(g) and H2O(l). (b) Without using thermochemical data, predict whether ¢G° for this reaction is more negative or less negative than ¢H°. 19.64 Sulfur dioxide reacts with strontium oxide as follows: SO21g2 + SrO1s2 ¡ SrSO31s2 (a) Without using thermochemical data, predict whether ¢G° for this reaction is more negative or less negative than ¢H°. (b) If you had only standard enthalpy data for this reaction, how would you go about making a rough estimate of the value of ¢G° at 298 K, using data from Appendix C on other substances? 19.65 Classify each of the following reactions as one of the four possible types summarized in Table 19.3: (a) N21g2 + 3 F21g2 ¡ 2 NF31g2 ¢H° = -249 kJ; ¢S° = - 278 J>K (b) N21g2 + 3 Cl 21g2 ¡ 2 NCl 31g2 ¢H° = 460 kJ; ¢S° = -275 J>K (c) N2F41g2 ¡ 2 NF21g2 ¢H° = 85 kJ; ¢S° = 198 J>K
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19.66 From the values given for ¢H° and ¢S°, calculate ¢G° for each of the following reactions at 298 K. If the reaction is not spontaneous under standard conditions at 298 K, at what temperature (if any) would the reaction become spontaneous? (a) 2 PbS1s2 + 3 O21g2 ¡ 2 PbO1s2 + 2 SO21g2 ¢H° = -844 kJ; ¢S° = -165 J>K (b) 2 POCl 31g2 ¡ 2 PCl 31g2 + O21g2 ¢H° = 572 kJ; ¢S° = 179 J>K 19.67 A particular constant-pressure reaction is spontaneous at 390 K. The enthalpy change for the reaction is +23.7 kJ. What can you conclude about the sign and magnitude of ¢S for the reaction? 19.68 A certain constant-pressure reaction is nonspontaneous at 45 °C. The entropy change for the reaction is 72 J>K. What can you conclude about the sign and magnitude of ¢H? 19.69 For a particular reaction, ¢H = - 32 kJ and ¢S = - 98 J>K. Assume that ¢H and ¢S do not vary with temperature. (a) At what temperature will the reaction have ¢G = 0? (b) If T is increased from that in part (a), will the reaction be spontaneous or nonspontaneous? 19.70 Reactions in which a substance decomposes by losing CO are called decarbonylation reactions. The decarbonylation of acetic acid proceeds as follows: CH3COOH1l2 ¡ CH3OH1g2 + CO1g2 By using data from Appendix C, calculate the minimum temperature at which this process will be spontaneous under standard conditions. Assume that ¢H° and ¢S° do not vary with temperature. 19.71 Consider the following reaction between oxides of nitrogen: NO21g2 + N2O1g2 ¡ 3 NO1g2 (a) Use data in Appendix C to predict how ¢G° for the reaction varies with increasing temperature. (b) Calculate ¢G° at 800 K, assuming that ¢H° and ¢S° do not change with temperature. Under standard conditions is the reaction spontaneous at 800 K? (c) Calculate ¢G° at 1000 K. Is the reaction spontaneous under standard conditions at this temperature?
19.72 Methanol (CH3OH) can be made by the controlled oxidation of methane: CH41g2 + 12 O21g2 ¡ CH3OH1g2 (a) Use data in Appendix C to calculate ¢H° and ¢S° for this reaction. (b) How is ¢G° for the reaction expected to vary with increasing temperature? (c) Calculate ¢G° at 298 K. Under standard conditions, is the reaction spontaneous at this temperature? (d) Is there a temperature at which the reaction would be at equilibrium under standard conditions and that is low enough so that the compounds involved are likely to be stable? 19.73 (a) Use data in Appendix C to estimate the boiling point of benzene, C6H6(l). (b) Use a reference source, such as the CRC Handbook of Chemistry and Physics, to find the experimental boiling point of benzene. How do you explain any deviation between your answer in part (a) and the experimental value? 19.74 (a) Using data in Appendix C, estimate the temperature at which the free-energy change for the transformation from I2(s) to I2(g) is zero. What assumptions must you make in arriving at this estimate? (b) Use a reference source, such as Web Elements (www.webelements.com), to find the experimental melting and boiling points of I2. (c) Which of the values in part (b) is closer to the value you obtained in part (a)? Can you explain why this is so? 19.75 Acetylene gas, C2H2(g), is used in welding. (a) Write a balanced equation for the combustion of acetylene gas to CO2(g) and H2O(l). (b) How much heat is produced in burning 1 mol of C2H2 under standard conditions if both reactants and products are brought to 298 K? (c) What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction? 19.76 The fuel in high-efficiency natural gas vehicles consists primarily of methane (CH4). (a) How much heat is produced in burning 1 mol of CH4(g) under standard conditions if reactants and products are brought to 298 K and H2O(l) is formed? (b) What is the maximum amount of useful work that can be accomplished under standard conditions by this system?
FREE ENERGY AND EQUILIBRIUM (section 19.7) 19.77 Explain qualitatively how ¢G changes for each of the following reactions as the partial pressure of O2 is increased: (a) 2 CO1g2 + O21g2 ¡ 2 CO21g2 (b) 2 H2O21l2 ¡ 2 H2O1l2 + O21g2 (c) 2 KClO31s2 ¡ 2 KCl1s2 + 3 O21g2
19.81 Use data from Appendix C to calculate the equilibrium constant, K, at 298 K for each of the following reactions: (a) H21g2 + I21g2 Δ 2 Hl1g2 (b) C2H5OH1g2 Δ C2H41g2 + H2O1g2 (c) 3 C2H21g2 Δ C6H61g2
19.78 Indicate whether ¢G increases, decreases, or does not change when the partial pressure of H2 is increased in each of the following reactions: (a) N21g2 + 3 H21g2 ¡ 2 NH31g2 (b) 2 HBr1g2 ¡ H21g2 + Br21g2 (c) 2 H21g2 + C2H21g2 ¡ C2H61g2
19.82 Using data from Appendix C, write the equilibrium-constant expression and calculate the value of the equilibrium constant for these reactions at 298 K: (a) NaHCO31s2 Δ NaOH1s2 + CO21g2 (b) 2 HBr1g2 + Cl 21g2 Δ 2 HCl1g2 + Br21g2 (c) 2 SO21g2 + O21g2 Δ 2 SO31g2
19.79 Consider the reaction 2 NO21g2 ¡ N2O41g2. (a) Using data from Appendix C, calculate ¢G° at 298 K. (b) Calculate ¢G at 298 K if the partial pressures of NO2 and N2O4 are 0.40 atm and 1.60 atm, respectively.
19.83 Consider the decomposition of barium carbonate:
19.80 Consider the reaction 3 CH41g2 ¡ C3H81g2 + 2 H21g2. (a) Using data from Appendix C, calculate ¢G° at 298 K. (b) Calculate ¢G at 298 K if the reaction mixture consists of 40.0 atm of CH4, 0.0100 atm of C3H8(g), and 0.0180 atm of H2.
BaCO31s2 Δ BaO1s2 + CO21g2 Using data from Appendix C, calculate the equilibrium pressure of CO2 at (a) 298 K and (b) 1100 K. 19.84 Consider the reaction PbCO31s2 Δ PbO1s2 + CO21g2 Using data in Appendix C, calculate the equilibrium pressure of CO2 in the system at (a) 400 °C and (b) 180 °C.
Additional Exercises 19.85 The value of Ka for nitrous acid (HNO2) at 25 °C is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to Ka. (b) By using the value of Ka, calculate ¢G° for the dissociation of nitrous acid in aqueous solution. (c) What is the value of ¢G at equilibrium? (d) What is the value of ¢G when 3H + 4 = 5.0 * 10 - 2 M, 3NO2 - 4 = 6.0 * 10 - 4 M, and 3HNO24 = 0.20 M?
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19.86 The Kb for methylamine (CH3NH2) at 25 °C is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to Kb. (b) By using the value of Kb, calculate ¢G° for the equilibrium in part (a). (c) What is the value of ¢G at equilibrium? (d) What is the value of ¢G when 3H + 4 = 6.7 * 10- 9 M, 3CH3NH3 + 4 = 2.4 × 10-3 M, and 3CH3NH24 = 0.098 M?
ADDITIONAL EXERCISES 19.87 (a) Which of the thermodynamic quantities T, E, q, w, and S are state functions? (b) Which depend on the path taken from one state to another? (c) How many reversible paths are there between two states of a system? (d) For a reversible isothermal process, write an expression for ¢E in terms of q and w and an expression for ¢S in terms of q and T. 19.88 Indicate whether each of the following statements is true or false. If it is false, correct it. (a) The feasibility of manufacturing NH3 from N2 and H2 depends entirely on the value of ¢H for the process N21g2 + 3 H21g2 ¡ 2 NH31g2. (b) The reaction of Na(s) with Cl2(g) to form NaCl(s) is a spontaneous process. (c) A spontaneous process can in principle be conducted reversibly. (d) Spontaneous processes in general require that work be done to force them to proceed. (e) Spontaneous processes are those that are exothermic and that lead to a higher degree of order in the system. 19.89 For each of the following processes, indicate whether the signs of ¢S and ¢H are expected to be positive, negative, or about zero. (a) A solid sublimes. (b) The temperature of a sample of Co(s) is lowered from 60 °C to 25 °C. (c) Ethyl alcohol evaporates from a beaker. (d) A diatomic molecule dissociates into atoms. (e) A piece of charcoal is combusted to form CO2(g) and H2O(g). 19.90 The reaction 2 Mg1s2 + O21g2 ¡ 2 MgO1s2 is highly spontaneous and has a negative value for ¢S°. The second law of thermodynamics states that in any spontaneous process there is always an increase in the entropy of the universe. Is there an inconsistency between this reaction and the second law? [19.91] Suppose four gas molecules are placed in the left flask in Figure 19.6(a). Initially, the right flask is evacuated and the stopcock is closed. (a) After the stopcock is opened, how many different arrangements of the molecules are possible? (b) How many of the arrangements from part (a) have all the molecules in the left flask? (c) How does the answer to part (b) explain the spontaneous expansion of the gas? [19.92] Consider a system that consists of two standard playing dice, with the state of the system defined by the sum of the values shown on the top faces. (a) The two arrangements of top faces shown here can be viewed as two possible microstates of the system. Explain. (b) To which state does each microstate correspond? (c) How many possible states are there for the system? (d) Which state or states have the highest entropy? Explain. (e) Which state or states have the lowest entropy? Explain.
19.93 Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of ¢S for this solution process? [19.94] A standard air conditioner involves a refrigerant that is typically now a fluorinated hydrocarbon, such as CH2F2. An air-conditioner refrigerant has the property that it readily vaporizes at atmospheric pressure and is easily compressed to its liquid phase under increased pressure. The operation of an air conditioner can be thought of as a closed system made up of the refrigerant going through the two stages shown here (the air circulation is not shown in this diagram). Expansion chamber Liquid
Vapor Expansion (low pressure) Compression chamber
Liquid
Vapor Compression (high pressure)
During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (a) What is the sign of q for the expansion? (b) What is the sign of q for the compression? (c) In a central air-conditioning system, one chamber is inside the home and the other is outside. Which chamber is where, and why? (d) Imagine that a sample of liquid refrigerant undergoes expansion followed by compression, so that it is back to its original state. Would you expect that to be a reversible process? (e) Suppose that a house and its exterior are both initially at 31 °C. Some time after the air conditioner is turned on, the house is cooled to 24 °C. Is this process spontaneous or nonspontaneous? [19.95] Trouton’s rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about 88 J>mol-K. (a) Estimate the normal boiling point of ° for Br2 using data from bromine, Br2, by determining ¢H vap ° remains constant with temAppendix C. Assume that ¢H vap perature and that Trouton’s rule holds. (b) Look up the normal boiling point of Br2 in a chemistry handbook or at the WebElements Web site (www.webelements.com). [19.96] For the majority of the compounds listed in Appendix C, the value of ¢G f° is more positive (or less negative) than the value of ¢H f°. (a) Explain this observation, using NH3(g), CCl4(l), and KNO3(s) as examples. (b) An exception to this observation is CO(g). Explain the trend in the ¢H f° and ¢G f° values for this molecule.
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CHAPTER 19
Chemical Thermodynamics
19.97 Consider the following three reactions: (i) Ti1s2 + 2 Cl 21g2 ¡ TiCl 41g2 (ii) C2H61g2 + 7 Cl 21g2 ¡ 2 CCl 41g2 + 6 HCl1g2 (iii) BaO1s2 + CO21g2 ¡ BaCO31s2 (a) For each of the reactions, use data in Appendix C to calculate ¢H°, ¢G°, and ¢S° at 25 °C. (b) Which of these reactions are spontaneous under standard conditions at 25 °C? (c) For each of the reactions, predict the manner in which the change in free energy varies with an increase in temperature. 19.98 Using the data in Appendix C and given the pressures listed, calculate ¢G° for each of the following reactions: (a) N21g2 + 3 H21g2 ¡ 2 NH31g2 PN2 = 2.6 atm, PH2 = 5.9 atm, PNH3 = 1.2 atm (b) 2 N2H41g2 + 2 NO21g2 ¡ 3 N21g2 + 4 H2O1g2 PN2H4 = PNO2 = 5.0 * 10-2 atm, PN2 = 0.5 atm, PH2O = 0.3 atm (c) N2H41g2 ¡ N21g2 + 2 H21g2 PN2H4 = 0.5 atm, PN2 = 1.5 atm, PH2 = 2.5 atm 19.99 (a) For each of the following reactions, predict the sign of ¢H° and ¢S° and discuss briefly how these factors determine the magnitude of K. (b) Based on your general chemical knowledge, predict which of these reactions will have K 7 0. (c) In each case indicate whether K should increase or decrease with increasing temperature. (i) 2 Mg1s2 + O21g2 Δ 2 MgO1s2 (ii) 2 KI1s2 Δ 2 K1g2 + I21g2 (iii) Na21g2 Δ 2 Na1g2 (iv) 2 V2O51s2 Δ 4 V1s2 + 5 O21g2 19.100 Acetic acid can be manufactured by combining methanol with carbon monoxide, an example of a carbonylation reaction: CH3OH1l2 + CO1g2 ¡ CH3COOH1l2 (a) Calculate the equilibrium constant for the reaction at 25 °C. (b) Industrially, this reaction is run at temperatures above 25 °C. Will an increase in temperature produce an increase or decrease in the mole fraction of acetic acid at equilibrium? Why are elevated temperatures used? (c) At what temperature will this reaction have an equilibrium constant equal to 1? (You may assume that ¢H° and ¢S° are temperature independent, and you may ignore any phase changes that might occur.) 19.101 The oxidation of glucose (C6H12O6) in body tissue produces CO2 and H2O. In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol (C2H5OH) and CO2. (a) Using data given in Appendix C, compare the equilibrium constants for the following reactions: C6H12O6 1s2 + 6 O21g2 Δ 6 CO21g2 + 6 H2O1l2 C6H12O6 1s2 Δ 2 C2 H5 OH1l2 + 2 CO21g2 (b) Compare the maximum work that can be obtained from these processes under standard conditions. [19.102] The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane (C2H6), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: 2 CH41g2 ¡ C2H61g2 + H21g2 In practice, this reaction is carried out in the presence of oxygen: 2 CH41g2 + 12 O21g2 ¡ C2H61g2 + H2O1g2 (a) Using the data in Appendix C, calculate K for these reactions at 25 °C and 500 °C. (b) Is the difference in ¢G° for the two reactions due primarily to the enthalpy term 1¢H2 or the
entropy term 1-T¢S2? (c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the “Chemistry and Life” box in Section 19.7. (d) The reaction of CH4 and O2 to form C2H6 and H2O must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction? [19.103] Cells use the hydrolysis of adenosine triphosphate (ATP) as a source of energy (Figure 19.19). The conversion of ATP to ADP has a standard free-energy change of -30.5 kJ>mol. If all the free energy from the metabolism of glucose, C6H12O6 1s2 + 6 O21g2 ¡ 6 CO21g2 + 6 H2O1l2 goes into the conversion of ADP to ATP, how many moles of ATP can be produced for each mole of glucose? [19.104] The potassium-ion concentration in blood plasma is about 5.0 * 10-3 M, whereas the concentration in muscle-cell fluid is much greater (0.15 M). The plasma and intracellular fluid are separated by the cell membrane, which we assume is permeable only to K + . (a) What is ¢G for the transfer of 1 mol of K + from blood plasma to the cellular fluid at body temperature 37 °C? (b) What is the minimum amount of work that must be used to transfer this K + ? [19.105] The relationship between the temperature of a reaction, its standard enthalpy change, and the equilibrium constant at that temperature can be expressed as the following linear equation: ln K =
- ¢H° + constant RT
(a) Explain how this equation can be used to determine ¢H° experimentally from the equilibrium constants at several different temperatures. (b) Derive the preceding equation using relationships given in this chapter. To what is the constant equal? [19.106] One way to derive Equation 19.3 depends on the observation that at constant T the number of ways, W, of arranging m ideal-gas particles in a volume V is proportional to the volume raised to the m power: W r Vm Use this relationship and Boltzmann’s relationship between entropy and number of arrangements (Equation 19.5) to derive the equation for the entropy change for the isothermal expansion or compression of n moles of an ideal gas. [19.107] About 86% of the world’s electrical energy is produced by using steam turbines, a form of heat engine. In his analysis of an ideal heat engine, Sadi Carnot concluded that the maximum possible efficiency is defined by the total work that could be done by the engine, divided by the quantity of heat available to do the work (for example, from hot steam produced by combustion of a fuel such as coal or methane). This efficiency is given by the ratio 1Thigh - Tlow2>Thigh, where Thigh is the temperature of the heat going into the engine and Tlow is that of the heat leaving the engine. (a) What is the maximum possible efficiency of a heat engine operating between an input temperature of 700 K and an exit temperature of 288 K? (b) Why is it important that electrical power plants be located near bodies of relatively cool water? (c) Under what conditions could a heat engine operate at or near 100% efficiency? (d) It is often said that if the energy of combustion of a fuel such as methane were captured in an electrical fuel cell instead of by burning the fuel in a heat engine, a greater fraction of the energy could be put to useful work. Make a qualitative drawing like that in Figure 5.10 that illustrates the fact that in principle the fuel cell route will produce more useful work than the heat engine route from combustion of methane.
Integrative Exercises
825
INTEGRATIVE EXERCISES 19.108 Most liquids follow Trouton’s rule, which states that the molar entropy of vaporization lies in the range of 88 ; 5 J>mol-K. The normal boiling points and enthalpies of vaporization of several organic liquids are as follows:
Substance
Normal Boiling ¢Hvap Point (°C) (kJ/mol)
Acetone, (CH3)2CO Dimethyl ether, (CH3)2O Ethanol, C2H5OH Octane, C8H18 Pyridine, C5H5N
56.1 - 24.8 78.4 125.6 115.3
29.1 21.5 38.6 34.4 35.1
(a) Calculate ¢Svap for each of the liquids. Do all the liquids obey Trouton’s rule? (b) With reference to intermolecular forces (Section 11.2), can you explain any exceptions to the rule? (c) Would you expect water to obey Trouton’s rule? By using data in Appendix B, check the accuracy of your conclusion. (d) Chlorobenzene (C6H5Cl) boils at 131.8 °C. Use Trouton’s rule to estimate ¢Hvap for this substance. 19.109 In chemical kinetics the entropy of activation is the entropy change for the process in which the reactants reach the activated complex. The entropy of activation for bimolecular processes is usually negative. Explain this observation with reference to Figure 14.17. 19.110 The following processes were all discussed in Chapter 18, “Chemistry of the Environment.” Estimate whether the entropy of the system increases or decreases during each process: (a) photodissociation of O2(g), (b) formation of ozone from oxygen molecules and oxygen atoms, (c) diffusion of CFCs into the stratosphere, (d) desalination of water by reverse osmosis. 19.111 Carbon disulfide (CS2) is a toxic, highly flammable substance. The following thermodynamic data are available for CS2(l) and CS2(g) at 298 K:
CS2(l) CS2(g)
¢Hf° (kJ/mol)
¢Gf° (kJ/mol)
89.7 117.4
65.3 67.2
(a) Draw the Lewis structure of the molecule. What do you predict for the bond order of the C ¬ S bonds? (b) Use the VSEPR method to predict the structure of the CS2 molecule. (c) Liquid CS2 burns in O2 with a blue flame, forming CO2(g) and SO2(g). Write a balanced equation for this reaction. (d) Using the data in the preceding table and in Appendix C, calculate ¢H° and ¢G° for the reaction in part (c). Is the reaction exothermic? Is it spontaneous at 298 K? (e) Use the data in the table to calculate ¢S° at 298 K for the vaporization of CS2(l). Is the sign of ¢S° as you would expect for a vaporization? (f) Using data in the table and your answer to part (e), estimate the boiling point of CS2(l). Do you predict that the substance will be a liquid or a gas at 298 K and 1 atm? [19.112] The following data compare the standard enthalpies and free energies of formation of some crystalline ionic substances and aqueous solutions of the substances:
Substance
≤H f° (kJ/mol)
≤Gf° (kJ/mol)
AgNO3(s) AgNO3(aq) MgSO4(s) MgSO4(aq)
-124.4 -101.7 -1283.7 -1374.8
-33.4 -34.2 -1169.6 -1198.4
(a) Write the formation reaction for AgNO3(s). Based on this reaction, do you expect the entropy of the system to increase or decrease upon the formation of AgNO3(s)? (b) Use ¢H f° and ¢G f° of AgNO3(s) to determine the entropy change upon formation of the substance. Is your answer consistent with your reasoning in part (a)? (c) Is dissolving AgNO3 in water an exothermic or endothermic process? What about dissolving MgSO4 in water? (d) For both AgNO3 and MgSO4, use the data to calculate the entropy change when the solid is dissolved in water. (e) Discuss the results from part (d) with reference to material presented in this chapter and in the “A Closer Look” box on page 540. [19.113] Consider the following equilibrium: N2O41g2 Δ 2 NO21g2 Thermodynamic data on these gases are given in Appendix C. You may assume that ¢H° and ¢S° do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? (b) At what temperature will an equilibrium mixture of 1 atm total pressure contain twice as much NO2 as N2O4? (c) At what temperature will an equilibrium mixture of 10 atm total pressure contain twice as much NO2 as N2O4? (d) Rationalize the results from parts (b) and (c) by using Le Châtelier’s principle. [Section 15.7] [19.114] The reaction SO2 1g2 + 2 H2S1g2 Δ 3 S1s2 + 2 H2O1g2 is the basis of a suggested method for removal of SO2 from power-plant stack gases. The standard free energy of each substance is given in Appendix C. (a) What is the equilibrium constant for the reaction at 298 K? (b) In principle, is this reaction a feasible method of removing SO2? (c) If PSO2 = PH2S and the vapor pressure of water is 25 torr, calculate the equilibrium SO2 pressure in the system at 298 K. (d) Would you expect the process to be more or less effective at higher temperatures? 19.115 When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here:
Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain constant temperature? (c) Try this experiment: Stretch a rubber band and wait a moment. Then place the stretched rubber band on your upper lip, and let it return suddenly to its unstretched state (remember to keep holding on). What do you observe? Are your observations consistent with your answer to part (b)?
WHAT’S AHEAD 20.1 OXIDATION STATES AND OXIDATION-REDUCTION REACTIONS We review oxidation states and oxidation-reduction (redox) reactions.
20.2 BALANCING REDOX EQUATIONS We learn how to balance redox equations using the method of half-reactions.
20.3 VOLTAIC CELLS We consider voltaic cells, which produce electricity from spontaneous redox reactions. Solid electrodes serve as the
surfaces at which oxidation and reduction take place. The electrode where oxidation occurs is the anode, and the electrode where reduction occurs is the cathode.
20.4 CELL POTENTIALS UNDER STANDARD CONDITIONS We see that an important characteristic of a voltaic cell is its cell potential, which is the difference in the electrical potentials at the two electrodes and is measured in units of volts. Half-cell potentials are tabulated for reduction half-reactions under standard conditions (standard reduction potentials).
20.5 FREE ENERGY AND REDOX REACTIONS We relate the Gibbs free energy, ¢G°, to cell potential.
20
A VARIETY OF BATTERIES of different sizes, composition, and voltages.
20.6 CELL POTENTIALS UNDER NONSTANDARD CONDITIONS We calculate cell potentials under nonstandard conditions by using standard cell potentials and the Nernst equation.
20.8 CORROSION We discuss corrosion, a spontaneous electrochemical process involving metals.
20.9 ELECTROLYSIS 20.7 BATTERIES AND FUEL CELLS We describe batteries and fuel cells, which are commercially important energy sources that use electrochemical reactions.
Finally, we focus on nonspontaneous redox reactions, examining electrolytic cells, which use electricity to perform chemical reactions.
ELECTROCHEMISTRY amazing array of portable electronic gadgets, including cell phones, portable music players, laptop computers, and gaming devices. In the absence of batteries, however, our electronic gadgetry would be nothing more than extra weight. Thus, a variety of batteries of different sizes,
WE ARE SURROUNDED BY AN
compositions, and voltages have been developed, some of which are shown in the chapter-opening photograph. Considerable research is in progress to develop new batteries with more power, faster recharging ability, lighter weight, or cheaper price. At the heart of such development are the oxidation-reduction reactions that power batteries. As we discussed in Chapter 4, oxidation is the loss of electrons in a chemical reaction, and reduction is the gain of electrons. • (Section 4.4) Thus, oxidationreduction (redox) reactions occur when electrons are transferred from an atom that is oxidized to an atom that is reduced. Redox reactions are involved not only in the operation of batteries but also in a wide variety of important natural processes, including the rusting of iron, the browning of foods, and the respiration of animals. Electrochemistry is the study of the relationships between electricity and chemical reactions. It includes the study of both spontaneous and nonspontaneous processes.
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CHAPTER 20
Electrochemistry
|
20.1 OXIDATION STATES AND OXIDATIONREDUCTION REACTIONS We determine whether a given chemical reaction is an oxidation-reduction reaction by keeping track of the oxidation numbers (oxidation states) of the elements involved in the reaction. • (Section 4.4) This procedure identifies whether the oxidation state changes for any elements involved in the reaction. For example, consider the reaction that occurs spontaneously when zinc metal is added to a strong acid (쑼 FIGURE 20.1): Zn(s) + 2 H + (aq) ¡ Zn2 + (aq) + H2(g)
[20.1]
The chemical equation for this reaction can be written H reduced
Zn(s) 2 H(aq) 0
1
Zn2(aq) H2(g) 2
[20.2]
0
Zn oxidized
From the oxidation numbers below the equation, we see that the oxidation number of Zn changes from 0 to +2 and that of H changes from +1 to 0. Thus, this is an oxidationreduction reaction. Electrons are transferred from zinc atoms to hydrogen ions and, therefore, Zn is oxidized and H + is reduced.
GO FIGURE
Explain (a) the vigorous bubbling in the beaker on the right and (b) the formation of steam above that beaker.
ⴙ ⴚ 씰 FIGURE 20.1 Oxidation of zinc by hydrochloric acid.
Zn(s)
ⴚ
ⴚ
ⴙ
2 HCl(aq)
2ⴙ
ⴚ
ZnCl2(aq)
H2(g)
SECTION 20.1
Oxidation States and Oxidation-Reduction Reactions
In a reaction such as Equation 20.2, a clear transfer of electrons occurs. In some reactions, however, the oxidation numbers change, but we cannot say that any substance literally gains or loses electrons. For example, in the combustion of hydrogen gas, 2 H2(g) O2(g) 0
2 H2O(g) 1
0
[20.3]
2
hydrogen is oxidized from the 0 to the +1 oxidation state and oxygen is reduced from the 0 to the -2 oxidation state. Therefore, Equation 20.3 is an oxidation-reduction reaction. Water is not an ionic substance, however, and so there is not a complete transfer of electrons from hydrogen to oxygen as water is formed. Thus, keeping track of oxidation states is a convenient form of “bookkeeping,” but you should not generally equate the oxidation state of an atom with its actual charge in a chemical compound. •(Section 8.5 “A Closer Look: Oxidation Numbers, Formal Charges, and Actual Partial Charges”) GIVE IT SOME THOUGHT What are the oxidation numbers of the elements in the nitrite ion, NO 2 - ?
In any redox reaction, both oxidation and reduction must occur. If one substance is oxidized, another must be reduced. The substance that makes it possible for another substance to be oxidized is called either the oxidizing agent or the oxidant. The oxidizing agent acquires electrons from the other substance and so is itself reduced. A reducing agent, or reductant, is a substance that gives up electrons, thereby causing another substance to be reduced. The reducing agent is therefore oxidized in the process. In Equation 20.2, H + (aq), the species that is reduced, is the oxidizing agent and Zn(s), the species that is oxidized, is the reducing agent. SAMPLE EXERCISE 20.1
Identifying Oxidizing and Reducing Agents
The nickel-cadmium (nicad) battery uses the following redox reaction to generate electricity: Cd(s) + NiO2(s) + 2 H2O(l) ¡ Cd(OH)2(s) + Ni(OH)2(s) Identify the substances that are oxidized and reduced, and indicate which is the oxidizing agent and which is the reducing agent. SOLUTION Analyze We are given a redox equation and asked to identify the substance oxidized and the substance reduced and to label the oxidizing agent and the reducing agent. Plan First, we assign oxidation states, or numbers, to all the atoms and determine which elements change oxidation state. Second, we apply the definitions of oxidation and reduction. Solve Cd( s) NiO2( s) 2H2O(l) 0
4 2
1 2
Cd(OH)2( s) Ni(OH)2( s)
2 2 1
2 2 1
The oxidation state of Cd increases from 0 to +2, and that of Ni decreases from +4 to +2. Thus, the Cd atom is oxidized (loses electrons) and is the reducing agent. The oxidation state of Ni decreases as NiO2 is converted into Ni(OH)2. Thus, NiO2 is reduced (gains electrons) and is the oxidizing agent. Comment A common mnemonic for remembering oxidation and reduction is “LEO the lion says GER”: losing electrons is oxidation; gaining electrons is reduction. PRACTICE EXERCISE Identify the oxidizing and reducing agents in the reaction 2 H2O(l) + Al(s) + MnO4 - (aq) ¡ Al(OH)4 - (aq) + MnO2(s) Answer: Al(s) is the reducing agent; MnO4 - (aq) is the oxidizing agent.
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CHAPTER 20
Electrochemistry
20.2 | BALANCING REDOX EQUATIONS Whenever we balance a chemical equation, we must obey the law of conservation of mass: The amount of each element must be the same on both sides of the equation. (Atoms are neither created nor destroyed in any chemical reaction.) As we balance oxidation-reduction reactions, there is an additional requirement: The gains and losses of electrons must be balanced. If a substance loses a certain number of electrons during a reaction, another substance must gain that same number of electrons. (Electrons are neither created nor destroyed in any chemical reaction.) In many simple chemical equations, such as Equation 20.2, balancing the electrons is handled “automatically” in the sense that we can balance the equation without explicitly considering the transfer of electrons. Many redox equations are more complex than Equation 20.2, however, and cannot be balanced easily without taking into account the number of electrons lost and gained. In this section we examine the method of halfreactions, a systematic procedure for balancing redox equations.
Half-Reactions Although oxidation and reduction must take place simultaneously, it is often convenient to consider them as separate processes. For example, the oxidation of Sn2 + by Fe 3 + , Sn2 + (aq) + 2 Fe 3 + (aq) ¡ Sn4 + (aq) + 2 Fe 2 + (aq) can be considered as consisting of two processes: oxidation of Sn2 + and reduction of Fe 3 + : Sn2 + (aq) ¡ Sn4 + (aq) + 2 e -
Oxidation: Reduction: 2 Fe
3+
(aq) + 2 e
-
¡ 2 Fe
2+
(aq)
[20.4] [20.5]
Notice that electrons are shown as products in the oxidation process and as reactants in the reduction process. Equations that show either oxidation or reduction alone, such as Equations 20.4 and 20.5, are called half-reactions. In the overall redox reaction, the number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction. When this condition is met and each half-reaction is balanced, the electrons on the two sides cancel when the two half-reactions are added to give the balanced oxidation-reduction equation.
Balancing Equations by the Method of Half-Reactions
(a)
“Other” atoms
(b)
O
(c)
H
(d)
e–
In using the half-reaction method, we usually begin with a “skeleton” ionic equation showing only the substances undergoing oxidation and reduction. In such cases, we usually do not need to assign oxidation numbers unless we are unsure whether the reaction involves oxidation-reduction. We will find that H + (for acidic solutions), OH - (for basic solutions), and H2O are often involved as reactants or products in redox reactions. Unless H + , OH - , or H2O is being oxidized or reduced, these species do not appear in the skeleton Balance atoms equation. Their presence, however, can be deduced as we balance the equation. other than H, O For balancing a redox reaction that occurs in acidic aqueous solution, the procedure is as follows: 1. Divide the equation into one oxidation half-reaction and one reduction halfBalance O reaction. 2. Balance each half-reaction. (a) First, balance elements other than H and O. (b) Next, balance O atoms by adding H2O as needed. Balance H (c) Then balance H atoms by adding H + as needed. (d) Finally, balance charge by adding e - as needed. This specific sequence (a)–(d) is important, and it is summarized in the diagram in the margin. At this point, you can check whether the number of electrons in each Balance electrons half-reaction corresponds to the changes in oxidation state.
SECTION 20.2
Balancing Redox Equations
831
3. Multiply half-reactions by integers as needed to make the number of electrons lost in the oxidation half-reaction equal the number of electrons gained in the reduction half-reaction. 4. Add half-reactions and, if possible, simplify by canceling species appearing on both sides of the combined equation. 5. Check to make sure that atoms and charges are balanced. As an example, let’s consider the reaction between permanganate ion (MnO4 - ) and oxalate ion (C2O42 - ) in acidic aqueous solution (쑼 FIGURE 20.2). When MnO4 - is added to an acidified solution of C2O42 - , the deep purple color of the MnO4 - ion fades, bubbles of CO2 form, and the solution takes on the pale pink color of Mn2 + . We can therefore write the skeleton equation as MnO4 - (aq) + C2O42 - (aq) ¡ Mn2 + (aq) + CO2(aq)
[20.6]
Experiments show that H + is consumed and H2O is produced in the reaction. We will see that their involvement in the reaction is deduced in the course of balancing the equation. To complete and balance Equation 20.6, we first write the two half-reactions (step 1). One half-reaction must have Mn on both sides of the arrow, and the other must have C on both sides of the arrow: MnO4 - (aq) ¡ Mn2 + (aq) C2O42 - (aq) ¡ CO2(g) We next complete and balance each half-reaction. First, we balance all the atoms except H and O (step 2a). In the permanganate half-reaction, we have one manganese atom on each side of the equation and so need to do nothing. In the oxalate halfreaction, we add a coefficient 2 on the right to balance the two carbons on the left: MnO4 - (aq) ¡ Mn2 + (aq) C2O42 - (aq) ¡ 2 CO2(g) Next we balance O (step 2b). The permanganate half-reaction has four oxygens on the left and none on the right; therefore we need four H2O molecules on the right to balance the oxygen atoms: MnO4 - (aq) ¡ Mn2 + (aq) + 4 H2O(l)
GO FIGURE
Which species is reduced in this reaction? Which species is the reducing agent? MnO4(aq)
The purple color of MnO4 disappears immediately as reaction with C2O42 occurs At end point, purple color of MnO4 remains because all C2O42 consumed C2O42(aq)
(a)
(b)
씱 FIGURE 20.2 Titration of an acidic solution of Na2C2O4 with KMnO 4(aq).
832
CHAPTER 20
Electrochemistry
The eight hydrogen atoms now in the products must be balanced by adding 8 H + to the reactants (step 2c): 8 H + (aq) + MnO4 - (aq) ¡ Mn2 + (aq) + 4 H2O(l) There are now equal numbers of each type of atom on the two sides of the equation, but the charge still needs to be balanced. The charge of the reactants is 8(1+) +1(1-) = 7 + , and that of the products is 1(2 +) +4(0) = 2+ . To balance the charge, we add five electrons to the reactant side (step 2d): 5 e - + 8 H + (aq) + MnO4 - (aq) ¡ Mn2 + (aq) + 4 H2O(l) We can use oxidation states to check our result. In this half-reaction Mn goes from the +7 oxidation state in MnO4 - to the +2 oxidation state of Mn2 + . Therefore, each Mn atom gains five electrons, in agreement with our balanced half-reaction. In the oxalate half-reaction, we have C and O balanced (step 2a). We balance the charge (step 2d) by adding two electrons to the products: C2O42 - (aq) ¡ 2 CO2(g) + 2 e We can check this result using oxidation states. Carbon goes from the +3 oxidation state in C2O42 - to the +4 oxidation state in CO2. Thus, each C atom loses one electron; therefore, the two C atoms in C2O42 - lose two electrons, in agreement with our balanced half-reaction. Now we multiply each half-reaction by an appropriate integer so that the number of electrons gained in one half-reaction equals the number of electrons lost in the other (step 3). We multiply the MnO4 - half-reaction by 2 and the C2O42 - half-reaction by 5: 10 e - + 16 H + (aq) + 2 MnO4 - (aq) ¡ 2 Mn2 + (aq) + 8 H2O(l) 5 C2O42 - (aq) ¡ 10 CO2(g) + 10 e 16 H + (aq) + 2 MnO4 - (aq) + 5 C2O42 - (aq) ¡ 2 Mn2 + (aq) + 8 H2O(l) + 10 CO2(g) The balanced equation is the sum of the balanced half-reactions (step 4). Note that the electrons on the reactant and product sides of the equation cancel each other. We check the balanced equation by counting atoms and charges (step 5): There are 16 H, 2 Mn, 28 O, 10 C, and a net charge of 4+ on each side of the equation, confirming that the equation is correctly balanced. GIVE IT SOME THOUGHT Do free electrons appear anywhere in the balanced equation for a redox reaction?
SAMPLE EXERCISE 20.2
Balancing Redox Equations in Acidic Solution
Complete and balance this equation by the method of half-reactions: Cr2O72 - (aq) + Cl - (aq) ¡ Cr 3 + (aq) + Cl 2(g)
(acidic solution)
SOLUTION Analyze We are given an incomplete, unbalanced (skeleton) equation for a redox reaction occurring in acidic solution and asked to complete and balance it. Plan We use the half-reaction procedure we just learned. Solve Step 1: We divide the equation into two half-reactions:
Cr2O72 - (aq) ¡ Cr 3 + (aq) Cl - (aq) ¡ Cl 2(g)
SECTION 20.2
Balancing Redox Equations
Step 2: We balance each half-reaction. In the first halfreaction the presence of one Cr2O72 - among the reactants requires two Cr 3 + among the products. The seven oxygen atoms in Cr2O72 - are balanced by adding seven H2O to the products. The 14 hydrogen atoms in 7 H2O are then balanced by adding 14 H + to the reactants:
14 H + (aq) + Cr2O72 - (aq) ¡ 2 Cr 3 + (aq) + 7 H2O(l)
We then balance the charge by adding electrons to the left side of the equation so that the total charge is the same on the two sides:
6 e - + 14 H + (aq) + Cr2O72 - (aq) ¡ 2 Cr 3 + (aq) + 7 H2O(l)
833
We can check this result by looking at the oxidation state changes. Each chromium atom goes from +6 to +3, gaining three electrons; therefore, the two Cr atoms in Cr2O72 - gain six electrons, in agreement with our half-reaction. In the second half-reaction, two Cl - are required to balance one Cl2:
2 Cl - (aq) ¡ Cl 2(g)
We add two electrons to the right side to attain charge balance:
2 Cl - (aq) ¡ Cl 2(g) + 2 e -
This result agrees with the oxidation state changes. Each chlorine atom goes from -1 to 0, losing one electron; therefore, the two chlorine atoms lose two electrons. Step 3: We equalize the number of electrons transferred in the two half-reactions. To do so, we multiply the Cl half-reaction by 3 so that the number of electrons gained in the Cr half-reaction (6) equals the number lost in the Cl half-reaction, allowing the electrons to cancel when the half-reactions are added:
6 Cl - (aq) ¡ 3 Cl 2(g) + 6 e -
Step 4: The equations are added to give the balanced equation:
14 H + (aq) + Cr2O72 - (aq) + 6 Cl - (aq) ¡ 2 Cr 3 + (aq) + 7 H2O(l) + 3 Cl 2(g)
Step 5: There are equal numbers of atoms of each kind on the two sides of the equation (14 H, 2 Cr, 7 O, 6 Cl). In addition, the charge is the same on the two sides (6 +). Thus, the equation is balanced. PRACTICE EXERCISE Complete and balance the following equations using the method of half-reactions. Both reactions occur in acidic solution. (a) Cu(s) + NO3 - (aq) ¡ Cu2 + (aq) + NO2(g) (b) Mn2 + (aq) + NaBiO3(s) ¡ Bi3 + (aq) + MnO4 - (aq) Answers: (a) Cu(s) + 4 H + (aq) + 2 NO3 - (aq) ¡ Cu2 + (aq) + 2 NO2(g) + 2 H2O(l)
(b) 2 Mn2 + (aq) + 5 NaBiO3(s) + 14 H + (aq) ¡ 2 MnO4 - (aq) + 5 Bi3 + (aq) + 5 Na + (aq) + 7 H2O(l)
Balancing Equations for Reactions Occurring in Basic Solution If a redox reaction occurs in basic solution, the equation must be balanced by using OH - and H2O rather than H + and H2O. One approach is to first balance the halfreactions as if they occurred in acidic solution and then count the number of H+ in each half-reaction and add the same number of OH - to each side of the half-reaction. This way, the reaction is mass-balanced because you are adding the same thing to both sides. In essence, what you are doing is “neutralizing” the protons to form water (H + + OH - ¡ H2O) on the side containing H + , and the other side ends up with the OH - . The resulting water molecules can be canceled as needed.
834
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Electrochemistry
SAMPLE EXERCISE 20.3
Balancing Redox Equations in Basic Solution
Complete and balance this equation for a redox reaction that takes place in basic solution: CN - (aq) + MnO4 - (aq) ¡ CNO - (aq) + MnO2(s)
(basic solution)
SOLUTION Analyze We are given an incomplete equation for a basic redox reaction and asked to balance it. Plan We go through the first steps of our procedure as if the reaction were occurring in acidic solution. We then add the appropriate number of OH - ions to each side of the equation, combining H + and OH - to form H2O. We complete the process by simplifying the equation. Solve Step 1: We write the incomplete, unbalanced half-reactions: Step 2: We balance each halfreaction as if it took place in acidic solution:
CN - (aq) ¡ CNO - (aq) MnO4 - (aq) ¡ MnO2(s) CN - (aq) + H2O(l) ¡ CNO - (aq) + 2 H + (aq) + 2 e 3 e - + 4 H + (aq) + MnO4 - (aq) ¡ MnO2(s) + 2 H2O(l)
Now we must take into account that the reaction occurs in basic solution, adding OH - to both sides of both half-reactions to neutralize H + :
3e
We “neutralize” H + and OH - by forming H2O when they are on the same side of either halfreaction:
CN - (aq) + H2O(l) + 2 OH - (aq) ¡ CNO - (aq) + 2 H2O(l) + 2 e 3 e - + 4 H2O(l) + MnO4 - (aq) ¡ MnO2(s) + 2 H2O(l) + 4 OH - (aq)
Next, we cancel water molecules that appear as both reactants and products:
CN - (aq) + H2O(l) + 2 OH - (aq) ¡ CNO - (aq) + 2 H + (aq) + 2 e - + 2 OH - (aq)
3e
-
-
+ 4 H (aq) + MnO4 - (aq) + 4 OH - (aq) ¡ MnO2(s) + 2 H2O(l) + 4 OH - (aq) +
CN - (aq) + 2 OH - (aq) ¡ CNO - (aq) + H2O(l) + 2 e + 2 H2O(l) + MnO4 - (aq) ¡ MnO2(s) + 4 OH - (aq)
Both half-reactions are now balanced. You can check the atoms and the overall charge. Step 3: We multiply the cyanide half-reaction by 3, which gives 6 electrons on the product side, and multiply the permanganate half-reaction by 2, which gives 6 electrons on the reactant side:
6 e-
Step 4: We add the two halfreactions together and simplify by canceling species that appear as both reactants and products:
3 CN - (aq) + H2O(l) + 2 MnO4 - (aq) ¡ 3 CNO - (aq) + 2 MnO2(s) + 2 OH - (aq)
3 CN - (aq) + 6 OH - (aq) ¡ 3 CNO - (aq) + 3 H2O(l) + 6 e + 4 H2O(l) + 2 MnO4 - (aq) ¡ 2 MnO2(s) + 8 OH - (aq)
Step 5: Check that the atoms and charges are balanced. There are 3 C, 3 N, 2 H, 9 O, 2 Mn, and a charge of 5 - on both sides of the equation. Comment It is important to remember that this procedure doesn’t imply that H+ ions are involved in the chemical reaction. Recall that in aqueous solutions at 20 °C, Kw = [H+][OH-] = 1.0 * 10-14. Thus, [H+ ] is very small in this basic solution. • (Section 16.3) PRACTICE EXERCISE Complete and balance the following equations for oxidation-reduction reactions that occur in basic solution: (a) NO2 - (aq) + Al(s) ¡ NH3(aq) + Al(OH)4 - (aq) (b) Cr(OH)3(s) + ClO - (aq) ¡ CrO42 - (aq) + Cl 2(g) Answers: (a) NO2 - (aq) + 2 Al(s) + 5 H2O(l) + OH - (aq) ¡ NH3(aq) + 2 Al(OH)4 - (aq)
(b) 2 Cr(OH)3(s) + 6 ClO - (aq) ¡ 2 CrO42 - (aq) + 3 Cl 2(g) + 2 OH - (aq) + 2 H2O(l)
SECTION 20.3
Voltaic Cells
835
GO FIGURE
Why does the intensity of the blue solution color lessen as the reaction proceeds?
Cu2 ions in solution
Atoms in Zn strip Electrons move from Zn to Cu2
2 e
Zn2 ion Cu atom
Zn oxidized
Cu2 reduced
Zn(s) Cu2(aq)
Zn2(aq) Cu(s)
쑿 FIGURE 20.3 A spontaneous oxidation-reduction reaction involving zinc and copper.
20.3 | VOLTAIC CELLS The energy released in a spontaneous redox reaction can be used to perform electrical work. This task is accomplished through a voltaic (or galvanic) cell, a device in which the transfer of electrons takes place through an external pathway rather than directly between reactants present in the same reaction vessel. One such spontaneous reaction occurs when a strip of zinc is placed in contact with a solution containing Cu2+. As the reaction proceeds, the blue color of Cu2 + (aq) ions fades, and copper metal deposits on the zinc. At the same time, the zinc begins to dissolve. These transformations, shown in 쑿 FIGURE 20.3, are summarized by the equation Zn(s) + Cu2 + (aq) ¡ Zn2 + (aq) + Cu(s)
GO FIGURE
Which metal, Cu or Zn, is oxidized in this voltaic cell? Zn electrode in 1 M ZnSO4 solution Cu electrode in 1 M CuSO4 solution
[20.7]
씰 FIGURE 20.4
shows a voltaic cell that uses the redox reaction given in Equation 20.7. Although the setup in Figure 20.4 is more complex than that in Figure 20.3, the reaction is the same in both cases. The significant difference is that in the voltaic cell the Zn metal and Cu2 + (aq) are not in direct contact with each other. Instead, Zn metal is in contact with Zn2 + (aq) in one compartment, and Cu metal is in contact with Cu2 + (aq) in the other compartment. Consequently, Cu2+ reduction can occur only by the flow of electrons through an external circuit, namely, a wire connecting the Zn and Cu strips. Electrons flowing through a wire and ions moving in solution both constitute an electrical current. This flow of electrical charge can be used to accomplish electrical work.
Solutions in contact with each other through porous glass disc 쑿 FIGURE 20.4 A Cu-Zn voltaic cell based on the reaction in Equation 20.7.
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Electrochemistry
The two solid metals connected by the external circuit are called electrodes. By definition, the electrode at which oxidation occurs is the anode and the electrode at which reduction occurs is the cathode.* The electrodes can be made of materials that participate in the reaction, as in the present example. Over the course of the reaction, the Zn electrode gradually disappears and the copper electrode gains mass. More typically, the electrodes are made of a conducting material, such as platinum or graphite, that does not gain or lose mass during the reaction but serves as a surface at which electrons are transferred. Each compartment of a voltaic cell is called a half-cell. One half-cell is the site of the oxidation half-reaction, and the other is the site of the reduction half-reaction. In our present example, Zn is oxidized and Cu2 + is reduced: Zn(s) ¡ Zn2 + (aq) + 2 e -
Anode (oxidation half-reaction)
Cu2 + (aq) + 2 e - ¡ Cu(s)
Cathode (reduction half-reaction)
Electrons become available as zinc metal is oxidized at the anode. They flow through the external circuit to the cathode, where they are consumed as Cu2 + (aq) is reduced. Because Zn(s) is oxidized in the cell, the zinc electrode loses mass, and the concentration of the Zn2 + solution increases as the cell operates. At the same time, the Cu electrode gains mass, and the Cu2 + solution becomes less concentrated as Cu2 + is reduced to Cu(s). For a voltaic cell to work, the solutions in the two half-cells must remain electrically neutral. As Zn is oxidized in the anode half-cell, Zn2 + ions enter the solution, upsetting the initial Zn2 + >SO42 - charge balance. To keep the solution electrically neutral, theremust be some means for Zn2 + cations to migrate out of the anode half-cell and for anions to migrate in. Similarly, the reduction of Cu2 + at the cathode removes these cations from the solution, leaving an excess of SO42 - anions in that half-cell. To maintain electrical neutrality, some of these anions must migrate out of the cathode half-cell, and positive ions must migrate in. In fact, no measurable electron flow occurs between electrodes unless a means is provided for ions to migrate through the solution from one half-cell to the other, thereby completing the circuit. In Figure 20.4, a porous glass disc separating the two half-cells allows ions to migrate and maintain the electrical neutrality of the solutions. In 쑼 FIGURE 20.5, a salt bridge GO FIGURE
How is electrical balance maintained in the left beaker as Zn2 + ions are formed at the anode? e
Anions migrate toward anode
1.10 V Voltmeter
e Cations migrate toward cathode
Zn
Cu NO3
Na
Salt bridge (allows ion migration)
Anode ()
Zn2
Cathode ()
Cu2
2e
Cu2 2
Zn
Zn(s)
Zn2(aq) 2 e
Cu2(aq) 2 e
Zn
Cu(s) Cu 2e
쑿 FIGURE 20.5 A voltaic cell that uses a salt bridge to complete the electrical circuit. *To help remember these definitions, note that anode and oxidation both begin with a vowel, and cathode and reduction both begin with a consonant.
SECTION 20.3
serves this purpose. The salt bridge consists of a U-shaped tube containing an electrolyte solution, such as NaNO3(aq), whose ions will not react with other ions in the voltaic cell or with the electrodes. The electrolyte is often incorporated into a paste or gel so that the electrolyte solution does not pour out when the U-tube is inverted. As oxidation and reduction proceed at the electrodes, ions from the salt bridge migrate into the two half-cells—cations migrating to the cathode half-cell and anions migrating to the anode half-cell—to neutralize charge in the half-cell solutions. Whichever device is used to allow ions to migrate between half-cells, anions always migrate toward the anode and cations toward the cathode. 씰 FIGURE 20.6 summarizes the various relationships in a voltaic cell. Notice in particular that electrons flow from the anode through the external circuit to the cathode. Because of this directional flow, the anode in a voltaic cell is labeled with a negative sign and the cathode is labeled with a positive sign. We can envision the electrons as being attracted to the positive cathode from the negative anode through the external circuit. SAMPLE EXERCISE 20.4
Describing a Voltaic Cell
The oxidation-reduction reaction Cr2O72 - (aq) + 14 H + (aq) + 6 I - (aq) ¡ 2 Cr 3 + (aq) + 3 I2(s) + 7 H2O(l) is spontaneous. A solution containing K2Cr2O7 and H2SO4 is poured into one beaker, and a solution of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution (such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron migration, the direction of ion migration, and the signs of the electrodes. SOLUTION Analyze We are given the equation for a spontaneous reaction that takes place in a voltaic cell and a description of how the cell is constructed. We are asked to write the half-reactions occurring at the anode and at the cathode, as well as the directions of electron and ion movements and the signs assigned to the electrodes. Plan Our first step is to divide the chemical equation into half-reactions so that we can identify the oxidation and the reduction processes. We then use the definitions of anode and cathode and the other terminology summarized in Figure 20.6. Solve In one half-reaction, Cr2O72 - (aq) is converted into Cr 3 + (aq). Starting with these ions and then completing and balancing the half-reaction, we have Cr2O72 - (aq) + 14 H + (aq) + 6 e - ¡ 2 Cr 3 + (aq) + 7 H2O(l) In the other half-reaction, I - (aq) is converted to I2(s): 6 I - (aq) ¡ 3 I2(s) + 6 e Now we can use the summary in Figure 20.6 to help us describe the voltaic cell. The first halfreaction is the reduction process (electrons on the reactant side of the equation). By definition, the reduction process occurs at the cathode. The second half-reaction is the oxidation process (electrons on the product side of the equation), which occurs at the anode. The I - ions are the source of electrons, and the Cr2O72 - ions accept the electrons. Hence, the electrons flow through the external circuit from the electrode immersed in the KI solution (the anode) to the electrode immersed in the K2Cr2O7 – H2SO4 solution (the cathode). The electrodes themselves do not react in any way; they merely provide a means of transferring electrons from or to the solutions. The cations move through the solutions toward the cathode, and the anions move toward the anode. The anode (from which the electrons move) is the negative electrode, and the cathode (toward which the electrons move) is the positive electrode. PRACTICE EXERCISE The two half-reactions in a voltaic cell are Zn(s) ¡ Zn2 + (aq) + 2 e ClO3 - (aq) + 6 H + (aq) + 6 e - ¡ Cl - (aq) + 3 H2O(l) (a) Indicate which reaction occurs at the anode and which at the cathode. (b) Which electrode is consumed in the cell reaction? (c) Which electrode is positive? Answers: (a) The first reaction occurs at the anode and the second reaction at the cathode. (b) The anode (Zn) is consumed in the cell reaction. (c) The cathode is positive.
Voltaic Cells
837
Electron flow e Anode ()
Voltmeter Porous barrier or salt bridge
e Cathode ()
Anions Cations Anode half-cell, oxidation occurs
Cathode half-cell, reduction occurs
쑿 FIGURE 20.6 Summary of reactions occurring in a voltaic cell. The half-cells can be separated by either a porous glass disc (as in Figure 20.4) or by a salt bridge (as in Figure 20.5).
838
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Electrochemistry
|
20.4 CELL POTENTIALS UNDER STANDARD CONDITIONS Why do electrons transfer spontaneously from a Zn atom to a Cu2 + ion, either directly as in Figure 20.3 or through an external circuit as in Figure 20.4? In a simple sense, we can compare the electron flow to the flow of water in a waterfall (쑼 FIGURE 20.7). Water flows spontaneously over a waterfall because of a difference in potential energy between the top of the falls and the bottom. • (Section 5.1) In a similar fashion, electrons flow spontaneously through an external circuit from the anode of a voltaic cell to the cathode because of a difference in potential energy. The potential energy of electrons is higher in the anode than in the cathode. Thus, electrons flow spontaneously toward the electrode with the more positive electrical potential. The difference in potential energy per electrical charge (the potential difference) between two electrodes is measured in volts. One volt (V) is the potential difference required to impart 1 joule (J) of energy to a charge of 1 coulomb (C): 1V = 1
J C
Recall that one electron has a charge of 1.60 * 10 - 19 C. • (Section 2.2) The potential difference between the two electrodes of a voltaic cell is called the cell potential, denoted Ecell. Because the potential difference provides the driving force that pushes electrons through the external circuit, we also call it the electromotive (“causing electron motion”) force, or emf. Because Ecell is measured in volts, it is also commonly called the voltage of the cell. The cell potential of any voltaic cell is positive. The magnitude of the cell potential depends on the reactions that occur at the cathode and anode, the concentrations of reactants and products, and the temperature, which we will assume to be 25 °C unless otherwise noted. In this section we focus on cells that are operated at 25 °C under standard conditions. Recall from Table 19.2 that standard conditions include 1 M concentrations for reactants and products in solution and 1 atm pressure for gaseous reactants and products. The cell potential under standard conditions is called either the ° . For the Zn-Cu voltaic cell standard cell potential or standard emf and is denoted E cell in Figure 20.5, for example, the standard cell potential at 25 °C is +1.10 V: Zn(s) + Cu2 + (aq, 1 M) ¡ Zn2 + (aq, 1 M) + Cu(s)
° = +1.10 V E cell
Recall that the superscript ° indicates standard-state conditions. • (Section 5.7)
High potential energy
Anode
Flow of electrons
Cathode 씰 FIGURE 20.7 Water analogy for electron flow.
Low potential energy
SECTION 20.4
Cell Potentials under Standard Conditions
839
GIVE IT SOME THOUGHT ° = +0.85 V at 25 °C, is the redox reaction of If a standard cell potential is E cell the cell spontaneous?
Standard Reduction Potentials ° , depends on the particular cathode and The standard cell potential of a voltaic cell, E cell anode half-cells. We could, in principle, tabulate the standard cell potentials for all possible cathode/anode combinations. However, it is not necessary to undertake this arduous task. Rather, we can assign a standard potential to each half-cell and then use ° . The cell potential is the difference between these half-cell potentials to determine E cell two half-cell potentials. By convention, the potential associated with each electrode is chosen to be the potential for reduction at that electrode. Thus, standard half-cell potentials are tabulated for reduction reactions, which means they are standard reduction ° . The standard cell potential, E cell ° , is the standard reduction potentials, denoted E red ° (cathode), minus the standard reduction potenpotential of the cathode reaction, E red ° (anode): tial of the anode reaction, E red ° = E red ° (cathode) - E red ° (anode) E cell
[20.8]
It is not possible to measure the standard reduction potential of a half-reaction directly. If we assign a standard reduction potential to a certain reference half-reaction, however, we can then determine the standard reduction potentials of other halfreactions relative to that reference value. The reference half-reaction is the reduction of H + (aq) to H2(g) under standard conditions, which is assigned a standard reduction potential of exactly 0 V: 2 H + (aq, 1 M) + 2 e - ¡ H2(g, 1 atm)
° = 0V E red
[20.9]
An electrode designed to produce this half-reaction is called a standard hydrogen electrode (SHE). An SHE consists of a platinum wire connected to a piece of platinum foil covered with finely divided platinum that serves as an inert surface for the reaction (쑼 FIGURE 20.8). The SHE allows the platinum to be in contact with both 1 M H + (aq) and a stream of hydrogen gas at 1 atm. The SHE can operate as either the anode or cathode of a cell, depending on the nature of the other electrode. 씰 FIGURE 20.9 shows a voltaic cell using an SHE. The spontaneous reaction is the one shown in Figure 20.1, namely, oxidation of Zn and reduction of H + : Zn(s) + 2 H + (aq) ¡ Zn2 + (aq) + H2(g)
Pt atom
H ion
H2 molecule
e e
1 atm H2(g)
Reduction SHE as cathode (H reduced to H2)
Pt wire Pt 1 M H
H2 molecule
H ion
e e
Oxidation SHE as anode (H2 oxidized to H)
씱 FIGURE 20.8 The standard hydrogen electrode (SHE) is used as a reference electrode.
840
CHAPTER 20
Electrochemistry
GO FIGURE
Why does Naⴙ migrate into the cathode half-cell as the cell reaction proceeds? e
e
0.76 V Voltmeter
Zn anode
NO3
Na H2(g)
씰 FIGURE 20.9 A voltaic cell using a standard hydrogen electrode (SHE). The anode half-cell is Zn metal in a Zn(NO3)2(aq) solution, and the cathode half-cell is the SHE in a HNO 3(aq) solution.
Anode half-cell Zn(s)
NO3 Zn2 NO3
NO3 H
Zn2(aq) 2 e
2 H(aq) 2 e
Cathode half-cell (standard hydrogen electrode, SHE)
H2(g)
When the cell is operated under standard conditions, the cell potential is +0.76 V. By ° = 0.76 V), the defined standard reduction potenusing the standard cell potential (E cell ° = 0 V) and Equation 20.8, we can determine the standard reduction tial of H + (E red potential for the Zn2 + >Zn half-reaction: ° = E red ° (cathode) - E red ° (anode) E cell ° (anode) +0.76 V = 0 V - E red ° (anode) = -0.76 V E red
Thus, a standard reduction potential of -0.76 V can be assigned to the reduction of Zn2 + to Zn: ° = -0.76 V E red Zn2 + (aq, 1 M) + 2 e - ¡ Zn(s) We write the reaction as a reduction even though the Zn reaction in Figure 20.9 is an oxidation. Whenever we assign an electrical potential to a half-reaction, we write the reaction as a reduction. Half-reactions, however, are reversible, being able to operate as either reductions or oxidations. Consequently, half-reactions are sometimes written using two arrows (Δ) between reactants and products, as in equilibrium reactions. The standard reduction potentials for other half-reactions can be determined in a fashion analogous to that used for the Zn2 + >Zn half-reaction. 씰 TABLE 20.1 lists some standard reduction potentials; a more complete list is found in Appendix E. These standard reduction potentials, often called half-cell potentials, can be combined to calculate ° values for a large variety of voltaic cells. E cell GIVE IT SOME THOUGHT For the half-reaction Cl2(g) + 2 e - ¡ 2 Cl - (aq), what are the standard conditions for the reactant and product?
Because electrical potential measures potential energy per electrical charge, standard reduction potentials are intensive properties. • (Section 1.3) In other words, if we increase the amount of substances in a redox reaction, we increase both the energy and the charges involved, but the ratio of energy (joules) to electrical charge (coulombs) remains constant (V = J>C). Thus, changing the stoichiometric coefficient in a half° for reaction does not affect the value of the standard reduction potential. For example, E red the reduction of 10 mol Zn2 + is the same as that for the reduction of 1 mol Zn2 + : 10 Zn2 + (aq, 1 M) + 20 e - ¡ 10 Zn(s)
° = -0.76 V E red
SECTION 20.4
Cell Potentials under Standard Conditions
TABLE 20.1 • Standard Reduction Potentials in Water at 25 °C ° (V) E red
Reduction Half-Reaction
+2.87
F2(g) + 2 e - ¡ 2 F - (aq)
+1.51
MnO4 - (aq) + 8 H + (aq) + 5 e - ¡ Mn2 + (aq) + 4 H2O(l)
+1.36
Cl 2(g) + 2 e - ¡ 2 Cl - (aq)
+1.33
Cr2O72 - (aq) + 14 H + (aq) + 6 e - ¡ 2 Cr 3 + (aq) + 7 H2O(l)
+1.23
O2(g) + 4 H + (aq) + 4 e - ¡ 2 H2O(l)
+1.06
Br2(l) + 2 e - ¡ 2 Br - (aq)
+0.96
NO3 - (aq) + 4 H + (aq) + 3 e - ¡ NO(g) + 2 H2O(l)
+0.80
Ag + (aq) + e - ¡ Ag(s)
+0.77
Fe 3 + (aq) + e - ¡ Fe 2 + (aq)
+0.68
O2(g) + 2 H + (aq) + 2 e - ¡ H2O2(aq)
+0.59
MnO4 - (aq) + 2 H2O(l) + 3 e - ¡ MnO2(s) + 4 OH - (aq)
+0.54
I2(s) + 2 e - ¡ 2 I - (aq)
+0.40
O2(g) + 2 H2O(l) + 4 e - ¡ 4 OH - (aq)
+0.34
Cu2 + (aq) + 2 e - ¡ Cu(s)
0 [defined]
2 H + (aq) + 2 e - ¡ H2(g)
-0.28
Ni2 + (aq) + 2 e - ¡ Ni(s)
-0.44
Fe 2 + (aq) + 2 e - ¡ Fe(s)
-0.76
Zn2 + (aq) + 2 e - ¡ Zn(s)
-0.83 -1.66
2 H2O(l) + 2 e - ¡ H2(g) + 2 OH - (aq) Al3 + (aq) + 3 e - ¡ Al(s)
-2.71
Na + (aq) + e - ¡ Na(s)
-3.05
Li + (aq) + e - ¡ Li(s)
SAMPLE EXERCISE 20.5 For the Zn-Cu
2+
° from E cell ° Calculating E red
voltaic cell shown in Figure 20.5, we have
Zn(s) + Cu2 + (aq, 1 M) ¡ Zn2 + (aq, 1 M) + Cu(s) 2+
Given that the standard reduction potential of Zn the reduction of Cu2 + to Cu:
° = 1.10 V E cell
° for to Zn(s) is -0.76 V, calculate the E red
Cu2 + (aq, 1 M) + 2 e - ¡ Cu(s) SOLUTION ° and E red ° for Zn2 + and asked to calculate E red ° for Cu2 + . Analyze We are given E cell ° for Plan In the voltaic cell, Zn is oxidized and is therefore the anode. Thus, the given E red ° (anode). Because Cu2 + is reduced, it is in the cathode half-cell. Thus, the unZn2 + is E red ° (cathode). Knowing E cell ° and E red ° (anode), we can known reduction potential for Cu2 + is E red ° (cathode). use Equation 20.8 to solve for E red
Solve ° = E red ° (cathode) - E red ° (anode) E cell ° (cathode) - (-0.76 V) 1.10 V = E red ° (cathode) = 1.10 V - 0.76 V = 0.34 V E red
Check This standard reduction potential agrees with the one listed in Table 20.1. ° 2 + = 0.34 V Comment: The standard reduction potential for Cu2 + can be represented as E Cu ° 2 + = -0.76 V. The subscript identifies the ion that is reduced in the and that for Zn2 + as E Zn reduction half-reaction.
841
842
Electrochemistry
CHAPTER 20
GO FIGURE ⴰ Ered
Given values for the two electrodes in a standard voltaic cell, how do you determine which electrode is the cathode? More positive
E°red (V)
Cathode (reduction) E°red(cathode)
PRACTICE EXERCISE The standard cell potential is 1.46 V for a voltaic cell based on the following half-reactions: In + (aq) ¡ In3 + (aq) + 2 e Br2(l) + 2 e - ¡ 2 Br - (aq) ° for the reduction of In3 + to In + . Using Table 20.1, calculate E red Answer: -0.40 V
SAMPLE EXERCISE 20.6
° from Ered ° Calculating Ecell
° for the voltaic cell described in Sample Exercise 20.4, which is Use Table 20.1 to calculate E cell based on the reaction
Cr2O72 - (aq) + 14 H + (aq) + 6 I - (aq) ¡ 2 Cr 3 + (aq) + 3 I2(s) + 7 H2O(l) E°cell E°red(cathode) E°red(anode)
Anode (oxidation)
SOLUTION Analyze We are given the equation for a redox reaction and asked to use data in Table 20.1 to calculate the standard cell potential for the associated voltaic cell. Plan Our first step is to identify the half-reactions that occur at the cathode and anode, which we did in Sample Exercise 20.4. Then we use Table 20.1 and Equation 20.8 to calculate the standard cell potential.
E°red(anode)
Solve The half-reactions are Cathode:
More negative
Anode:
쑿 FIGURE 20.10 Graphical representation of standard cell potential of a voltaic cell.
Cr2O72 - (aq) + 14 H + (aq) + 6 e - ¡ 2 Cr 3 + (aq) + 7 H2O(l) 6 I - (aq) ¡ 3 I2(s) + 6 e -
According to Table 20.1, the standard reduction potential for the reduction of Cr2O72 - to Cr 3 + is +1.33 V and the standard reduction potential for the reduction of I2 to I - (the reverse of the oxidation half-reaction) is +0.54 V. We use these values in Equation 20.8: ° = E red ° (cathode) - E red ° (anode) = 1.33 V - 0.54 V = 0.79 V E cell
Although we must multiply the iodide half-reaction by 3 to obtain a balanced equation, we do ° value by 3. As we have noted, the standard reduction potential is an intennot multiply the E red sive property and so is independent of the stoichiometric coefficients.
More positive
E°red (V)
0.34
0.76
Check The cell potential, 0.79 V, is a positive number. As noted earlier, a voltaic cell must have a positive potential.
Cu2 2 e
Cu Cathode
E°cell (0.34) (0.76) 1.10 V
Anode Zn
Zn
2
2 e
쑿 FIGURE 20.11 Half-cell potentials and standard cell potential for the Zn-Cu voltaic cell.
PRACTICE EXERCISE Using data in Table 20.1, calculate the standard emf for a cell that employs the overall cell reaction 2 Al(s) + 3 I2(s) ¡ 2 Al3 + (aq) + 6 I - (aq). Answer: 0.54 V - (-1.66 V) = 2.20 V
For each half-cell in a voltaic cell, the standard reduction potential provides a meas° , the greater ure of the tendency for reduction to occur: The more positive the value of E red the tendency for reduction under standard conditions. In any voltaic cell operating under ° value for the reaction at the cathode is more positive than standard conditions, the E red ° the E red value for the reaction at the anode. Thus, electrons flow spontaneously through ° to the electhe external circuit from the electrode with the more negative value of E red ° trode with the more positive value of E red. The fact that the standard cell potential is the difference between the standard reduction potentials of cathode and anode is illustrated graphically in 씱 FIGURE 20.10. ° value identifies the cathode, and the difference between the two The more positive E red ° standard reduction potentials is the standard cell potential. 씱 FIGURE 20.11 shows E red values for the two half-reactions in the Zn-Cu voltaic cell of Figure 20.5.
SECTION 20.4
SAMPLE EXERCISE 20.7
Cell Potentials under Standard Conditions
Determining Half-Reactions at Electrodes and Calculating Cell Potentials
A voltaic cell is based on the two standard half-reactions Cd2 + (aq) + 2 e - ¡ Cd(s) Sn2 + (aq) + 2 e - ¡ Sn(s) Use data in Appendix E to determine (a) which half-reaction occurs at the cathode and which occurs at the anode and (b) the standard cell potential. SOLUTION ° for two half-reactions. We then use these values first to Analyze We have to look up E red ° . determine the cathode and the anode and then to calculate the standard cell potential, E cell ° value, and the anode Plan The cathode will have the reduction with the more positive E red ° . To write the half-reaction at the anode, we reverse the halfwill have the less positive E red reaction written for the reduction, so that the half-reaction is written as an oxidation.
Solve ° (Cd2 + >Cd) = -0.403 V and E red ° (Sn2 + >Sn) = -0.136 V. (a) According to Appendix E, E red 2+ The standard reduction potential for Sn is more positive (less negative) than that for Cd2 + . Hence, the reduction of Sn2 + is the reaction that occurs at the cathode: Cathode:
Sn2 + (aq) + 2 e - ¡ Sn(s)
The anode reaction, therefore, is the loss of electrons by Cd: Anode:
Cd(s) ¡ Cd2 + (aq) + 2 e -
(b) The cell potential is given by the difference in the standard reduction potentials at the cathode and anode (Equation 20.8): ° = E red ° (cathode) - E red ° (anode) = (-0.136 V) - (-0.403 V) = 0.267 V E cell ° values of both half-reactions are negative; the negaNotice that it is unimportant that the E red tive values merely indicate how these reductions compare to the reference reaction, the reduction of H + (aq).
Check: The cell potential is positive, as it must be for a voltaic cell. PRACTICE EXERCISE A voltaic cell is based on a Co2 + >Co half-cell and an AgCl>Ag half-cell. (a) What half-reaction occurs at the anode? (b) What is the standard cell potential? Answers: (a) Co ¡ Co2 + + 2 e - ; (b) +0.499 V
Strengths of Oxidizing and Reducing Agents Table 20.1 lists half-reactions in order of decreasing tendency to undergo reduction. For ° . example, F2 is located at the top of the table, having the most positive value for E red Thus, F2 is the most easily reduced species in Table 20.1 and therefore the strongest oxidizing agent listed. Among the most frequently used oxidizing agents are the halogens, O2, and oxyanions such as MnO4 - , Cr2O72 - , and NO3 - , whose central atoms have high positive ° oxidation states. As seen in Table 20.1, all these species have large positive values of E red and therefore easily undergo reduction. The lower the tendency for a half-reaction to occur in one direction, the greater the tendency for it to occur in the opposite direction. Thus, the half-reaction with the most negative reduction potential in Table 20.1 is the one most easily reversed and run as an oxidation. Being at the bottom of Table 20.1, Li + (aq) is the most difficult species in the list to reduce and is therefore the poorest oxidizing agent listed. Although Li + (aq) has little tendency to gain electrons, the reverse reaction, oxidation of Li(s) to Li + (aq), is highly favorable. Thus, Li is the strongest reducing agent among the substances listed in Table 20.1. (Note that, because Table 20.1 lists half-reactions as reductions, only the
843
844
CHAPTER 20
Electrochemistry
substances on the reactant side of these equations can serve as oxidizing agents; only those on the product side can serve as reducing agents.) Commonly used reducing agents include H2 and the active metals, such as the alkali ° metals and the alkaline earth metals. Other metals whose cations have negative E red values—Zn and Fe, for example—are also used as reducing agents. Solutions of reducing agents are difficult to store for extended periods because of the ubiquitous presence of O2, a good oxidizing agent. The information contained in Table 20.1 is summarized graphically in 쑼 FIGURE 20.12. The reactants in half-reactions at the top of Table 20.1 are the most readily reduced species in the table and are therefore the strongest oxidizing agents. Thus, Figure 20.12 shows F2(g) as the strongest oxidizing agent (the position at the top of the red arrow). The products in half-reactions at the top of Table 20.1 are the most difficult to oxidize and are therefore the weakest reducing agents in the table. The reactants in half-reactions at the bottom of Table 20.1 are the most difficult to reduce and so are the weakest oxidizing agents. The products in half-reactions at the bottom of Table 20.1 are the most readily oxidized species in the table and so are the strongest reducing agents. This inverse relationship between oxidizing and reducing strength is similar to the inverse relationship between the strengths of conjugate acids and bases. • (Section 16.2 and Figure 16.3)
GO FIGURE
Why is a strong oxidizing agent easy to reduce? Most positive values of E˚red
Easiest to reduce, strongest oxidizing agent
Most difficult to oxidize, weakest reducing agent
씰 FIGURE 20.12 Relative strengths of oxidizing and reducing agents. The standard reduction potentials in Table 20.1 are related to the ability of substances to serve as oxidizing or reducing agents. Species on the left side of the half-reactions can act as oxidizing agents, and those on the right side can act as reducing agents.
Most difficult to reduce, weakest oxidizing agent
2 H(aq) 2 e
Li(aq) e
2 F(aq)
H2(g)
Li(s)
Most negative values of E˚red
Increasing strength of reducing agent
Increasing strength of oxidizing agent
F2(g) 2 e
Easiest to oxidize, strongest reducing agent
SECTION 20.5
SAMPLE EXERCISE 20.8
Free Energy and Redox Reactions
845
Determining Relative Strengths of Oxidizing Agents
Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents: NO3 - (aq), Ag + (aq), Cr2O72 - (aq). SOLUTION Analyze We are asked to rank the abilities of several ions to act as oxidizing agents. ° value), the stronger it is as Plan The more readily an ion is reduced (the more positive its E red an oxidizing agent.
Solve From Table 20.1, we have NO3 - (aq) + 4 H + (aq) + 3 e - ¡ NO(g) + 2 H2O(l) +
Ag (aq) + e Cr2O7
2-
+
(aq) + 14 H (aq) + 6 e
-
° = + 0.80 V E red
¡ Ag(s) ¡ 2 Cr
3+
° = + 0.96 V E red
° = + 1.33 V (aq) + 7 H2O(l) E red
Because the standard reduction potential of Cr2O72 + oxidizing agent of the three. The rank order is Ag
is the most positive, Cr2O72 - is the strongest 6 NO3 - 6 Cr2O72 - .
PRACTICE EXERCISE Using Table 20.1, rank the following species from the strongest to the weakest reducing agent: I - (aq), Fe(s), Al(s). Answer: Al(s) 7 Fe(s) 7 I - (aq)
20.5 | FREE ENERGY AND REDOX REACTIONS We have observed that voltaic cells use spontaneous redox reactions to produce a positive cell potential. We can use this fact together with half-cell potentials to decide whether a given redox reaction is spontaneous. In doing so, it is useful to make Equation 20.8 more general so that we can see how it pertains to general redox reactions, not just reactions in voltaic cells: ° (reduction process) - E red ° (oxidation process) E° = E red
[20.10]
In writing the equation this way, we have dropped the subscript “cell” to indicate that the calculated emf does not necessarily refer to a voltaic cell. Also, we have generalized the standard reduction potentials by using the general terms reduction and oxidation rather than the terms specific to voltaic cells, cathode and anode. We can now make a general statement about the spontaneity of a reaction and its associated emf, E: A positive value of E indicates a spontaneous process; a negative value of E indicates a nonspontaneous process. We use E to represent the emf under nonstandard conditions and E° to indicate the standard emf.
SAMPLE EXERCISE 20.9
Determining Spontaneity
Use Table 20.1 to determine whether the following reactions are spontaneous under standard conditions. (a) Cu(s) + 2 H + (aq) ¡ Cu2 + (aq) + H2(g) (b) Cl 2(g) + 2 I - (aq) ¡ 2 Cl - (aq) + I2(s) SOLUTION Analyze We are given two reactions and must determine whether each is spontaneous. Plan To determine whether a redox reaction is spontaneous under standard conditions, we first need to write its reduction and oxidation half-reactions. We can then use the standard reduction potentials and
Equation 20.10 to calculate the standard emf, E°, for the reaction. If a reaction is spontaneous, its standard emf must be a positive number.
846
CHAPTER 20
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Solve (a) For oxidation of Cu to Cu2 + and reduction of H + to H2, the half-reactions and standard reduction potentials are
Reduction: Oxidation:
Notice that for the oxidation, we use the standard reduction potential from Table 20.1 for the reduction of Cu2 + to Cu. We now calculate E° by using Equation 20.10:
° (reduction process) - E red ° (oxidation process) E° = E red = (0 V) - (0.34 V) = -0.34 V
Because E° is negative, the reaction is not spontaneous in the direction written. Copper metal does not react with acids in this fashion. The reverse reaction, however, is spontaneous and has a positive E° value: Thus, Cu2 + can be reduced by H2.
2 H + (aq) + 2 e - ¡ H2(g) Cu(s) ¡ Cu2 + (aq) + 2 e -
Cu2 + (aq) + H2(g) ¡ Cu(s) + 2 H + (aq)
E° = +0.34 V
Cl 2(g) + 2 e - ¡ 2 Cl - (aq) 2 I - (aq) ¡ I2(s) + 2 e -
(b) We follow a procedure analogous to that in (a):
Reduction: Oxidation:
In this case
E° = (1.36 V) - (0.54 V) = +0.82 V
° = 0V E red ° = +0.34 V E red
E°red = +1.36 V E°red = +0.54 V
Because the value of E° is positive, this reaction is spontaneous and could be used to build a voltaic cell. PRACTICE EXERCISE Using the standard reduction potentials listed in Appendix E, determine which of the following reactions are spontaneous under standard conditions: (a) I2(s) + 5 Cu2 + (aq) + 6 H2O(l) ¡ 2 IO3 - (aq) + 5 Cu(s) + 12 H + (aq) (b) Hg 2 + (aq) + 2 I - (aq) ¡ Hg(l) + I2(s) (c) H2SO3(aq) + 2 Mn(s) + 4 H + (aq) ¡ S(s) + 2 Mn2 + (aq) + 3 H2O(l) Answer: Reactions (b) and (c) are spontaneous.
We can use standard reduction potentials to understand the activity series of metals. •(Section 4.4) Recall that any metal in the activity series (Table 4.5) is oxidized by the ions of any metal below it. We can now recognize the origin of this rule based on standard reduction potentials. The activity series is based on the oxidation reactions of the metals, ordered from strongest reducing agent at the top to weakest reducing agent at the bottom. (Thus, the ordering is inverted relative to that in Table 20.1.) For example, nickel lies above silver in the activity series, making nickel the stronger reducing agent. Because a reducing agent is oxidized in any redox reaction, nickel is more easily oxidized than silver. In a mixture of nickel metal and silver cations, therefore, we expect a displacement reaction in which the silver ions are displaced in the solution by nickel ions: Ni(s) + 2 Ag + (aq) ¡ Ni2 + (aq) + 2 Ag(s) In this reaction Ni is oxidized and Ag + is reduced. Therefore, the standard emf for the reaction is ° (Ag +>Ag2 - E red ° (Ni2+>Ni2 E° = E red
= (+0.80 V) - (-0.28 V) = +1.08 V The positive value of E° indicates that the displacement of silver by nickel resulting from oxidation of Ni metal and reduction of Ag + is a spontaneous process. Remember that although we multiply the silver half-reaction by 2, the reduction potential is not multiplied. GIVE IT SOME THOUGHT Based on Table 4.5, which is the stronger reducing agent, Hg(l) or Pb(s)?
SECTION 20.5
Free Energy and Redox Reactions
847
Emf, Free Energy, and the Equilibrium Constant The change in the Gibbs free energy, ¢G, is a measure of the spontaneity of a process that occurs at constant temperature and pressure. • (Section 19.5) The emf, E, of a redox reaction also indicates whether the reaction is spontaneous. The relationship between emf and the free-energy change is ¢G = - nFE
[20.11]
In this equation, n is a positive number without units that represents the number of moles of electrons transferred according to the balanced equation for the reaction, and F is Faraday’s constant, named after Michael Faraday (씰 FIGURE 20.13): F = 96,485 C>mol = 96,485 J>V-mol Faraday’s constant is the quantity of electrical charge on 1 mol of electrons. The units of ¢G calculated with Equation 20.11 are J>mol. As in Equation 19.19, we use “per mole” to mean per mole of reaction as indicated by the coefficients in the balanced equation. • (Section 19.7) Because both n and F are positive numbers, a positive value of E in Equation 20.11 leads to a negative value of ¢G. Remember: A positive value of E and a negative value of ¢G both indicate a spontaneous reaction. When the reactants and products are all in their standard states, Equation 20.11 can be modified to relate ¢G° and E°: ¢G° = - nFE°
[20.12]
Because ¢G° is related to the equilibrium constant, K, for a reaction by the expression ¢G° = - RT ln K (Equation 19.20), we can relate E° to K by solving Equation 20.12 for E° and then substituting the Equation 19.20 expression for ¢G°: E° =
¢G° -RT ln K RT = = ln K -nF -nF nF
쑼 FIGURE 20.14 summarizes the relationships among
[20.13]
E°, ¢G°, and K.
쑿 FIGURE 20.13 Michael Faraday. Faraday (1791–1867) was born in England, a child of a poor blacksmith. At the age of 14 he was apprenticed to a bookbinder who gave him time to read and to attend lectures. In 1812 he became an assistant in Humphry Davy’s laboratory at the Royal Institution. He succeeded Davy as the most famous and influential scientist in England, making an amazing number of important discoveries, including his formulation of the quantitative relationships between electrical current and the extent of chemical reaction in electrochemical cells.
GO FIGURE
What does the variable n represent in the ≤G° and E° equations? G° nFE°
G°
G° RT ln K
E° E° RT ln K nF
K
쑿 FIGURE 20.14 Relationships of E°, ≤G°, and K. Any one of these important parameters can be used to calculate the other two. The signs of E° and ¢G° determine the direction that the reaction proceeds under standard conditions. The magnitude of K determines the relative amounts of reactants and products in an equilibrium mixture.
Using Standard Reduction Potentials to Calculate ≤G° and K
SAMPLE EXERCISE 20.10
(a) Use the standard reduction potentials in Table 20.1 to calculate the standard free-energy change, ¢G°, and the equilibrium constant, K, at 298 K for the reaction 4 Ag(s) + O2(g) + 4 H + (aq) ¡ 4 Ag + (aq) + 2 H2O(l) (b) Suppose the reaction in part (a) is written 2 Ag(s) +
1 2
O2(g) + 2 H + (aq) ¡ 2 Ag + (aq) + H2O(l)
What are the values of E°, ¢G°, and K when the reaction is written in this way?
848
CHAPTER 20
Electrochemistry
SOLUTION Analyze We are asked to determine ¢G° and K for a redox reaction, using standard reduction potentials. Plan We use the data in Table 20.1 and Equation 20.10 to determine E° for the reaction and then use E° in Equation 20.12 to calculate ¢G°. We can then use Equation 19.20, ¢G° = - RT ln K, to calculate K. RT Alternatively, we can calculate K using Equation 20.13, E° = ln K. nF Solve (a) We first calculate E° by breaking the equation into two half-reactions and ° values from Table 20.1 obtaining E red (or Appendix E):
° = + 1.23 V Reduction: O2(g) + 4 H + (aq) + 4 e - ¡ 2 H2O(l) E red ° = + 0.80 V Oxidation: 4 Ag(s) ¡ 4 Ag + (aq) + 4 e - E red
Even though the second half-reaction ° value directly has 4 Ag, we use the E red from Table 20.1 because emf is an intensive property. Using Equation 20.10, we have
E° = (1.23 V) - (0.80 V) = 0.43 V
The half-reactions show the transfer of four electrons. Thus, for this reaction n = 4. We now use Equation 20.12 to calculate ¢G°:
¢G° = - nFE° = - (4)(96,485 J>V-mol)(+ 0.43 V) = - 1.7 * 105 J/mol = - 170 kJ/mol
The positive value of E° leads to a negative value of ¢G°. The per mol part of the unit relates to the balanced equation, 4 Ag(s) + O2(g) + 4 H + (aq) ¡ 4 Ag + (aq) + 2 H2O(l). Thus, -170 kJ is associated with 4 mol Ag, 1 mol O2 and 4 mol H + , and so forth, corresponding to the coefficients in the balanced equation. Now we need to calculate the equilibrium constant, K, using ¢G° = RT ln K. Because ¢G° is a large negative number, which means the reaction is thermodynamically very favorable, we expect K to be large.
¢G° = - RT ln K - 1.7 * 105 J>mol = - (8.314 J>K mol) (298 K) ln K - 1.7 * 105 J>mol -(8.314 J>K mol)(298 K) ln K = 69 K = 9 * 1029 ln K =
K is indeed very large! This means that we expect silver metal to oxidize in acidic aqueous environments, in air, to Ag + . Notice that the emf calculated for the reaction was E° = 0.43 V, which is easy to measure. Directly measuring such a large equilibrium constant by measuring reactant and product concentrations at equilibrium, on the other hand, would be very difficult. (b) The overall equation is the same as that in part (a), multiplied by 12 . The half-reactions are
Reduction:
1 2
O2(g) + 2 H + (aq) + 2 e - ¡ H2O(l)
Oxidation:
+
2 Ag(s) ¡ 2 Ag (aq) + 2 e
° are the same as they The values of E red were in part (a); they are not changed by multiplying the half-reactions by 12. Thus, E° has the same value as in part (a):
E° = + 0.43 V
Notice, though, that the value of n has changed to n = 2, which is one-half the value in part (a). Thus, ¢G° is half as large as in part (a):
¢G° = - (2)(96,485 J>V-mol)(+0.43 V) = - 83 kJ>mol
° = + 1.23 V E red -
° = + 0.80 V E red
The value of ¢G° is half that in part (a) because the coefficients in the chemical equation are half those in (a). Now we can calculate K as before:
- 8.3 * 104 J>mol = - (8.314 J>K mol)(298 K) ln K K = 4 * 1014
Comment E° is an intensive quantity, so multiplying a chemical equation by a certain factor will not affect the value of E°. Multiplying an equation will change the value of n, however, and hence the value
of ¢G°. The change in free energy, in units of J>mol of reaction as written, is an extensive quantity. The equilibrium constant is also an extensive quantity.
SECTION 20.6
Cell Potentials under Nonstandard Conditions
849
PRACTICE EXERCISE For the reaction 3 Ni2 + (aq) + 2 Cr(OH)3(s) + 10 OH - (aq) ¡ 3 Ni(s) + 2 CrO42 - (aq) + 8 H2O(l) (a) What is the value of n? (b) Use the data in Appendix E to calculate ¢G°. (c) Calculate K at T = 298 K. Answers: (a) 6, (b) +87 kJ>mol, (c) K = 6 * 10 - 16
A CLOSER LOOK ELECTRICAL WORK For any spontaneous process, ¢G is a measure of the maximum useful work, wmax, that can be extracted from the process: ¢G = wmax. • (Section 19.5) Because ¢G = -nFE, the maximum useful electrical work obtainable from a voltaic cell is wmax = -nFEcell
[20.14]
Because cell emf, Ecell, is always positive for a voltaic cell, wmax is negative, indicating that work is done by a system on its surroundings, as we expect for a voltaic cell. • (Section 5.2) As Equation 20.14 shows, the more charge a voltaic cell moves through a circuit (that is, the larger nF is) and the larger the emf pushing the electrons through the circuit (that is, the larger Ecell is), the more work the cell can accomplish. In Sample Exercise 20.10, we calculated ¢G° = -170 kJ>mol for the reaction 4 Ag(s) + O2(g) + 4 H + (aq) ¡ 4 Ag + (aq) + 2 H2O(l). Thus, a voltaic cell utilizing this reaction could perform a maximum of 170 kJ of work in consuming 4 mol Ag, 1 mol O2, and 4 mol H + .
If a reaction is not spontaneous, ¢G is positive and E is negative. To force a nonspontaneous reaction to occur in an electrochemical cell, we need to apply an external potential, Eext, that exceeds ƒ Ecell ƒ . For example, if a nonspontaneous process has E = -0.9 V, then the external potential Eext must be greater than +0.9 V in order for the process to occur. We will examine such nonspontaneous processes in Section 20.9. Electrical work can be expressed in energy units of watts times time. The watt (W) is a unit of electrical power (that is, rate of energy expenditure): 1 W = 1 J>s Thus, a watt-second is a joule. The unit employed by electric utilities is the kilowatt-hour (kWh), which equals 3.6 * 106 J: 1 kWh = (1000 W)(1 hr) a
RELATED EXERCISES: 20.59, 20.60
|
20.6 CELL POTENTIALS UNDER NONSTANDARD CONDITIONS We have seen how to calculate the emf of a cell when the reactants and products are under standard conditions. As a voltaic cell is discharged, however, reactants are consumed and products are generated, so concentrations change. The emf progressively drops until E = 0, at which point we say the cell is “dead.” In this section we examine how the emf generated under nonstandard conditions can be calculated by using an equation first derived by Walther Nernst (1864–1941), a German chemist who established many of the theoretical foundations of electrochemistry.
The Nernst Equation The effect of concentration on cell emf can be obtained from the effect of concentration on free-energy change. •(Section 19.7) Recall that the free-energy change for any chemical reaction, ¢G, is related to the standard free-energy change for the reaction, ¢G°: ¢G = ¢G° + RT ln Q
[20.15]
The quantity Q is the reaction quotient, which has the form of the equilibrium-constant expression except that the concentrations are those that exist in the reaction mixture at a given moment. • (Section 15.6)
3600 s 1 J>s ba b = 3.6 * 106 J 1 hr 1W
850
CHAPTER 20
Electrochemistry
Substituting ¢G = -nFE (Equation 20.11) into Equation 20.15 gives -nFE = -nFE° + RT ln Q Solving this equation for E gives the Nernst equation: E = E° -
RT ln Q nF
[20.16]
This equation is customarily expressed in terms of the base-10 logarithm: E = E° -
2.303 RT log Q nF
[20.17]
At T = 298 K the quantity 2.303 RT>F equals 0.0592, with units of volts, and so the Nernst equation simplifies to E = E° -
0.0592 V log Q n
(T = 298 K)
[20.18]
We can use this equation to find the emf E produced by a cell under nonstandard conditions or to determine the concentration of a reactant or product by measuring E for the cell. For example, consider the following reaction: Zn(s) + Cu2 + (aq) ¡ Zn2 + (aq) + Cu(s) In this case n = 2 (two electrons are transferred from Zn to Cu2 + ), and the standard emf is +1.10 V (Section 20.4). Thus, at 298 K the Nernst equation gives E = 1.10 V -
[Zn2 + ] 0.0592 V log 2 [Cu2 + ]
[20.19]
Recall that pure solids are excluded from the expression for Q. • (Section 15.6) According to Equation 20.19, the emf increases as [Cu2 + ] increases and as [Zn2 + ] decreases. For example, when [Cu2 + ] is 5.0 M and [Zn2 + ] is 0.050 M, we have 0.0592 V 0.050 log a b 2 5.0 0.0592 V = 1.10 V (-2.00) = 1.16 V 2
E = 1.10 V -
Thus, increasing the concentration of reactant Cu2 + and decreasing the concentration of product Zn2 + relative to standard conditions increases the emf of the cell relative to standard conditions (E° = +1.10 V). The Nernst equation helps us understand why the emf of a voltaic cell drops as the cell discharges. As reactants are converted to products, the value of Q increases, so the value of E decreases, eventually reaching E = 0. Because ¢G = -nFE (Equation 20.11), it follows that ¢G = 0 when E = 0. Recall that a system is at equilibrium when ¢G = 0. • (Section 19.7) Thus, when E = 0, the cell reaction has reached equilibrium, and no net reaction occurs. In general, increasing the concentration of reactants or decreasing the concentration of products increases the driving force for the reaction, resulting in a higher emf. Conversely, decreasing the concentration of reactants or increasing the concentration of products causes the emf to decrease from its value under standard conditions. SAMPLE EXERCISE 20.11
Cell Potential under Nonstandard Conditions
Calculate the emf at 298 K generated by a voltaic cell in which the reaction is Cr2O72 - (aq) + 14 H + (aq) + 6 I - (aq) ¡ 2 Cr 3 + (aq) + 3 I2(s) + 7 H2O(l) when [Cr2O72 - ] = 2.0 M, [H + ] = 1.0 M, [I - ] = 1.0 M, and [Cr 3 + ] = 1.0 * 10 - 5 M. SOLUTION Analyze We are given a chemical equation for a voltaic cell and the concentrations of reactants and products under which it operates. We are asked to calculate the emf of the cell under these nonstandard conditions.
SECTION 20.6
Cell Potentials under Nonstandard Conditions
Plan To calculate the emf of a cell under nonstandard conditions, we use the Nernst equation in the form of Equation 20.18. Solve We calculate E° for the cell from standard reduction potentials (Table 20.1 or Appendix E). The standard emf for this reaction was calculated in Sample Exercise 20.6: E° = 0.79 V. As that exercise shows, six electrons are transferred from reducing agent to oxidizing agent, so n = 6. The reaction quotient, Q, is Q =
[Cr 3 + ]2 [Cr2O7
2-
+ 14
- 6
][H ] [I ]
=
(1.0 * 10 - 5 )2 (2.0)(1.0)14 (1.0)6
= 5.0 * 10 - 11
Using Equation 20.18, we have E = 0.79 V - a = 0.79 V - a
0.0592 V b log(5.0 * 10 - 11) 6 0.0592 V b (-10.30) 6
= 0.79 V + 0.10 V = 0.89 V Check This result is qualitatively what we expect: Because the concentration of Cr2O72 (a reactant) is greater than 1 M and the concentration of Cr 3 + (a product) is less than 1 M, the emf is greater than E°. Because Q is about 10 - 10, log Q is about -10. Thus, the correction to E° is about 0.06 * 10>6, which is 0.1, in agreement with the more detailed calculation. PRACTICE EXERCISE Calculate the emf generated by the cell described in the practice exercise accompanying Sample Exercise 20.6 when [Al3 + ] = 4.0 * 10 - 3 M and [I - ] = 0.010 M. Answer: E = +2.36 V SAMPLE EXERCISE 20.12
Calculating Concentrations in a Voltaic Cell
+
If the potential of a Zn-H cell (like that in Figure 20.9) is 0.45 V at 25 °C when [Zn2 + ] = 1.0 M and PH2 = 1.0 atm, what is the H + concentration? SOLUTION Analyze We are given a description of a voltaic cell, its emf, and the concentration of Zn2 + and the partial pressure of H2 (both products in the cell reaction). We are asked to calculate the concentration of H + , a reactant. Plan We write the equation for the cell reaction and use standard reduction potentials to calculate E° for the reaction. After determining the value of n from our reaction equation, we solve the Nernst equation, Equation 20.18, for Q. Finally, we use the equation for the cell reaction to write an expression for Q that contains [H + ] to determine [H + ]. Solve The cell reaction is
Zn(s) + 2 H + (aq) ¡ Zn2 + (aq) + H2(g)
The standard emf is
° (reduction) - E red ° (oxidation) E° = E red = 0 V - (-0.76 V) = +0.76 V n = 2 0.0592 V 0.45 V = 0.76 V log Q 2 Q = 1010.5 = 3 * 1010
Because each Zn atom loses two electrons, Using Equation 20.18, we can solve for Q:
Q has the form of the equilibrium constant for the reaction: Solving for 3H + 4, we have
[Zn2 + ]PH2
(1.0)(1.0) = = 3 * 1010 [H + ]2 [H + ]2 1.0 3H + 42 = = 3 * 10 - 11 3 * 1010 Q =
3H + 4 = 23 * 10 - 11 = 6 * 10 - 6 M Comment A voltaic cell whose cell reaction involves H + can be used to measure [H + ] or pH. A pH meter is a specially designed voltaic cell with a voltmeter calibrated to read pH directly. •(Section 16.4) PRACTICE EXERCISE What is the pH of the solution in the cathode half-cell in Figure 20.9 when PH2 = 1.0 atm, [Zn2 + ] in the anode half-cell is 0.10 M, and the cell emf is 0.542 V? Answer: pH = 4.23 (using data from Appendix E to obtain E° to three significant figures)
851
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Concentration Cells In the voltaic cells we have looked at thus far, the reactive species at the anode has been different from the reactive species at the cathode. Cell emf depends on concentration, however, so a voltaic cell can be constructed using the same species in both half-cells as long as the concentrations are different. A cell based solely on the emf generated because of a difference in a concentration is called a concentration cell. An example of a concentration cell is diagrammed in 쑼 FIGURE 20.15(a). One halfcell consists of a strip of nickel metal immersed in a 1.00 * 10 - 3 M solution of Ni2 + (aq). The other half-cell also has an Ni(s) electrode, but it is immersed in a 1.00 M solution of Ni2 + (aq). The two half-cells are connected by a salt bridge and by an external wire running through a voltmeter. The half-cell reactions are the reverse of each other: Ni(s) ¡ Ni2 + (aq) + 2 e Ni2 + (aq) + 2 e - ¡ Ni(s)
Anode: Cathode:
° = -0.28 V E red ° = -0.28 V E red
Although the standard emf for this cell is zero, E°cell = E°red (cathode) - E°red (anode) = (-0.28 V) - (-0.28 V) = 0 V the cell operates under nonstandard conditions because the concentration of Ni2 + (aq) is not 1 M in both half-cells. In fact, the cell operates until [Ni2 + ]anode = [Ni2 + ]cathode. Oxidation of Ni(s) occurs in the half-cell containing the more dilute solution, which means this is the anode of the cell. Reduction of Ni2 + (aq) occurs in the half-cell containing the more concentrated solution, making it the cathode. The overall cell reaction is therefore Anode: Cathode:
Ni(s) ¡ Ni2 + (aq, dilute) + 2 e Ni2 + (aq, concentrated) + 2 e - ¡ Ni(s)
Overall:
Ni2 + (aq, concentrated) ¡ Ni2 + (aq, dilute)
We can calculate the emf of a concentration cell by using the Nernst equation. For this particular cell, we see that n = 2. The expression for the reaction quotient for the overall reaction is Q = [Ni2 + ]dilute>[Ni2 + ]concentrated. Thus, the emf at 298 K is E = E° -
0.0592 V log Q n
[Ni2 + ]dilute 0.0592 V 1.00 * 10 - 3 M 0.0592 V log = log 2 2 1.00 M [Ni2 + ]concentrated = +0.0888 V = 0 -
GO FIGURE
Assuming that the solutions are made from Ni(NO3)2, how do the ions migrate as the cell operates? e Ni anode
0.0888 V
e
0.0000 V
Salt bridge Discharge
Ni cathode
[Ni2] 1.00 103 M [Ni2] 1.00 M (a) 쑿 FIGURE 20.15 Concentration cell based on the Ni2ⴙ-Ni cell reaction. (a) Concentrations of Ni2 + (aq) in the two half-cells are unequal, and the cell generates an electrical current. (b) The cell operates until [Ni2 + (aq)] is the same in the two half-cells, at which point the cell has reached equilibrium and is “dead.”
[Ni2] 0.5 M
[Ni2] 0.5 M (b)
SECTION 20.6
Cell Potentials under Nonstandard Conditions
853
This concentration cell generates an emf of nearly 0.09 V even though E° = 0. The difference in concentration provides the driving force for the cell. When the concentrations in the two half-cells become the same, Q = 1 and E = 0. The idea of generating a potential by a concentration difference is the basis for the operation of pH meters. It is also a critical aspect in biology. For example, nerve cells in the brain generate a potential across the cell membrane by having different concentrations of ions on the two sides of the membrane. The regulation of the heartbeat in mammals, as discussed in the following “Chemistry and Life” box, is another example of the importance of electrochemistry to living organisms.
CHEMISTRY AND LIFE
The human heart is a marvel of efficiency and dependability. In a typical day an adult’s heart pumps more than 7000 L of blood through the circulatory system, usually with no maintenance required beyond a sensible diet and lifestyle. We generally think of the heart as a mechanical device, a muscle that circulates blood via regularly spaced muscular contractions. However, more than two centuries ago, two pioneers in electricity, Luigi Galvani (1729–1787) and Alessandro Volta (1745–1827), discovered that the contractions of the heart are controlled by electrical phenomena, as are nerve impulses throughout the body. The pulses of electricity that cause the heart to beat result from a remarkable combination of electrochemistry and the properties of semipermeable membranes. • (Section 13.5) Cell walls are membranes with variable permeability with respect to a number of physiologically important ions (especially Na + , K + , and Ca2 + ). The concentrations of these ions are different for the fluids inside the cells (the intracellular fluid, or ICF) and outside the cells (the extracellular fluid, or ECF). In cardiac muscle cells, for example, the concentrations of K + in the ICF and ECF are typically about 135 millimolar (mM) and 4 mM, respectively. For Na + , however, the concentration difference between the ICF and ECF is opposite that for K + ; typically, [Na + ]ICF = 10 mM and [Na + ]ECF = 145 mM. The cell membrane is initially permeable to K + ions but is much less so to Na + and Ca2 + . The difference in concentration of K + ions between the ICF and ECF generates a concentration cell: Even though the same ions are present on both sides of the membrane, there is a potential difference between the two fluids that we can calculate using the Nernst equation with E° = 0. At physiological temperature (37 °C) the potential in millivolts for moving K + from the ECF to the ICF is
Time 쑿 FIGURE 20.16 Changes in electrical potential in the human heart. Variation of the electrical potential caused by changes of ion concentrations in the pacemaker cells of the heart.
in 쑿 FIGURE 20.16. The emf cycle determines the rate at which the heart beats. If the pacemaker cells malfunction because of disease or injury, an artificial pacemaker can be surgically implanted. The artificial pacemaker contains a small battery that generates the electrical pulses needed to trigger the contractions of the heart. During the late 1800s, scientists discovered that the electrical impulses that cause the contraction of the heart muscle are strong enough to be detected at the surface of the body. This observation formed the basis for electrocardiography, noninvasive monitoring of the heart by using a complex array of electrodes on the skin to measure voltage changes during heartbeats. A typical electrocardiogram is shown in 쑼 FIGURE 20.17. It is quite striking that, although the heart’s major function is the mechanical pumping of blood, it is most easily monitored by using the electrical impulses generated by tiny voltaic cells.
1 second
[K + ]ICF 2.30 RT log + nF [K ]ECF
= 0 - (61.5 mV) log a
135 mM b = -94 mV 4 mM
In essence, the interior of the cell and the ECF together serve as a voltaic cell. The negative sign for the potential indicates that work is required to move K + into the intracellular fluid. Changes in the relative concentrations of the ions in the ECF and ICF lead to changes in the emf of the voltaic cell. The cells of the heart that govern the rate of heart contraction are called the pacemaker cells. The membranes of the cells regulate the concentrations of ions in the ICF, allowing them to change in a systematic way. The concentration changes cause the emf to change in a cyclic fashion, as shown
emf
E = E° -
Electrical potential
HEARTBEATS AND ELECTROCARDIOGRAPHY
Time 쑿 FIGURE 20.17 A typical electrocardiogram. The printout records the electrical events monitored by electrodes attached to the body surface.
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SAMPLE EXERCISE 20.13
Determining pH Using a Concentration Cell
A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has PH2 = 1.00 atm and an unknown concentration of H + (aq). Electrode 2 is a standard hydrogen electrode (PH2 = 1.00 atm, [H + ] = 1.00 M). At 298 K the measured cell potential is 0.211 V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate [H + ] for the solution at electrode 1. What is the pH of the solution? SOLUTION Analyze We are given the potential of a concentration cell and the direction in which the current flows. We also have the concentrations or partial pressures of all reactants and products except for [H + ] in half-cell 1, which is our unknown. Plan We can use the Nernst equation to determine Q and then use Q to calculate the unknown concentra° = 0 V. tion. Because this is a concentration cell, E cell Solve Using the Nernst equation, we have
0.211 V = 0 -
0.0592 V log Q 2
log Q = -(0.211 V)a
2 b = -7.13 0.0592 V
Q = 10 - 7.13 = 7.4 * 10 - 8 Because electrons flow from electrode 1 to electrode 2, electrode 1 is the anode of the cell and electrode 2 is the cathode. The electrode reactions are therefore as follows, with the concentration of H + (aq) in electrode 1 represented with the unknown x:
Electrode: 2
+
2 H (aq; 1.00 M) + 2 e
Overall: Q =
Thus,
H2(g, 1.00 atm) ¡ 2 H + (aq, x M) + 2 e -
Electrode: 1
=
-
¡ H2(g, 1.00 atm)
+
° = 0V E red ° = 0V E red
+
2 H (aq; 1.00 M) ¡ 2 H (aq, x M)
[H + (aq, x M)]2 [H + (aq, 1.00M)]2 x2 = x 2 = 7.4 * 10 - 8 (1.00)2
x = 3H + 4 = 27.4 * 10 - 8 = 2.7 * 10 - 4 At electrode 1, therefore, the pH of the solution is
pH = - log[H + ] = -log(2.7 * 10 - 4) = 3.57
Comment The concentration of H + at electrode 1 is lower than that in electrode 2, which is why electrode 1 is the anode of the cell: The oxidation of H2 to H + (aq) increases [H + ] at electrode 1. PRACTICE EXERCISE A concentration cell is constructed with two Zn(s)– Zn2 + (aq) half-cells. In one half-cell [Zn2 + ] = 1.35 M, and in the other [Zn2 + ] = 3.75 * 10 - 4 M. (a) Which half-cell is the anode? (b) What is the emf of the cell? Answers: (a) the half-cell in which [Zn2 + ] = 3.75 * 10 - 4 M, (b) 0.105 V
20.7 | BATTERIES AND FUEL CELLS
1.5 V
3.0 V
1.5 V
A battery is a portable, self-contained electrochemical power source that consists of one or more voltaic cells. For example, the 1.5-V batteries used to power flashlights and many consumer electronic devices are single voltaic cells. Greater voltages can be achieved by using multiple cells, as in 12-V automotive batteries. When cells are connected in series (which means the cathode of one attached to the anode of another), the battery produces a voltage that is the sum of the voltages of the individual cells. Higher voltages can also be achieved by using multiple batteries in series (씱 FIGURE 20.18). Battery electrodes are marked following the convention of Figure 20.6—plus for cathode and minus for anode. 씱 FIGURE 20.18 Combining batteries. When batteries are connected in series, as in most flashlights, the total voltage is the sum of the individual voltages.
SECTION 20.7
Although any spontaneous redox reaction can serve as the basis for a voltaic cell, making a commercial battery that has specific performance characteristics can require considerable ingenuity. The substances oxidized at the anode and reduced by the cathode determine the voltage, and the usable life of the battery depends on the quantities of these substances packaged in the battery. Usually a barrier analogous to the porous barrier of Figure 20.6 separates the anode and cathode half-cells. Different applications require batteries with different properties. The battery required to start a car, for example, must be capable of delivering a large electrical current for a short time period, whereas the battery that powers a heart pacemaker must be very small and capable of delivering a small but steady current over an extended time period. Some batteries are primary cells, meaning they cannot be recharged and must be either discarded or recycled after the voltage drops to zero. A secondary cell can be recharged from an external power source after its voltage has dropped. As we consider some common batteries, notice how the principles we have discussed so far help us understand these important sources of portable electrical energy.
Batteries and Fuel Cells
GO FIGURE
What is the oxidation state of lead in the cathode of this battery?
Lead grid filled with spongy lead (anode)
H2SO4 electrolyte
Lead grid filled with PbO2 (cathode)
Lead-Acid Battery A 12-V lead-acid automotive battery consists of six voltaic cells in series, each producing 2 V. The cathode of each cell is lead dioxide (PbO2) packed on a lead grid (씰 FIGURE 20.19). The anode of each cell is lead. Both electrodes are immersed in sulfuric acid. The reactions that occur during discharge are Cathode:
855
쑿 FIGURE 20.19 A 12-V automotive lead-acid battery. Each anode/cathode pair in this schematic cutaway produces a voltage of about 2 V. Six pairs of electrodes are connected in series, producing 12 V.
PbO2(s) + HSO4 - (aq) + 3 H + (aq) + 2 e - ¡ PbSO4(s) + 2 H2O(l) Pb(s) + HSO4 - (aq) ¡ PbSO4(s) + H + (aq) + 2 e -
Anode:
PbO2(s) + Pb(s) + 2 HSO4 - (aq) + 2 H + (aq) ¡ 2 PbSO4(s) + 2 H2O(l)
[20.20]
The standard cell potential can be obtained from the standard reduction potentials in Appendix E: ° = E red ° (cathode) - E red ° (anode) = (+1.685 V) - (-0.356 V) = +2.041V E cell
The reactants Pb and PbO2 are the electrodes. Because these reactants are solids, there is no need to separate the cell into half-cells; the Pb and PbO2 cannot come into contact with each other unless one electrode touches another. To keep the electrodes from touching, wood or glass-fiber spacers are placed between them (Figure 20.19). Using a reaction whose reactants and products are solids has another benefit. Because solids are excluded from the reaction quotient Q, the relative amounts of Pb(s), PbO2(s), and PbSO4(s) have no effect on the voltage of the lead storage battery, helping the battery maintain a relatively constant voltage during discharge. The voltage does vary somewhat with use because the concentration of H2SO4 varies with the extent of discharge. As Equation 20.20 indicates, H2SO4 is consumed during the discharge. A major advantage of the lead-acid battery is that it can be recharged. During recharging, an external source of energy is used to reverse the direction of the cell reaction, regenerating Pb(s) and PbO2(s):
Separator
2 PbSO4(s) + 2 H2O(l) ¡ PbO2(s) + Pb(s) + 2 HSO4 - (aq) + 2 H + (aq) In an automobile the alternator provides the energy necessary for recharging the battery. Recharging is possible because PbSO4 formed during discharge adheres to the electrodes. As the external source forces electrons from one electrode to the other, the PbSO4 is converted to Pb at one electrode and to PbO2 at the other.
Alkaline Battery The most common primary (nonrechargeable) battery is the alkaline battery (씰 FIGURE 20.20). The anode is powdered zinc metal immobilized in a gel in contact with a concentrated solution of KOH (hence, the name alkaline battery). The cathode is a mixture of
Cathode (MnO2 Anode plus graphite) (Zn plus KOH) 쑿 FIGURE 20.20 Cutaway view of a miniature alkaline battery.
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MnO2(s) and graphite, separated from the anode by a porous fabric. The battery is sealed in a steel can to reduce the risk of any of the concentrated KOH escaping. The cell reactions are complex but can be approximately represented as follows: Cathode: Anode:
2 MnO2(s) + 2 H2O(l) + 2 e - ¡ 2 MnO(OH)(s) + 2 OH - (aq) Zn(s) + 2 OH - (aq) ¡ Zn(OH)2(s) + 2 e -
Nickel-Cadmium, Nickel-Metal-Hydride, and Lithium-Ion Batteries The tremendous growth in high-power-demand portable electronic devices in the last decade has increased the demand for lightweight, readily recharged batteries. One of the most common rechargeable batteries is the nickel-cadmium (nicad) battery. During discharge, cadmium metal is oxidized at the anode while nickel oxyhydroxide [NiO(OH)(s)] is reduced at the cathode: Cathode: Anode:
2 NiO(OH)(s) + 2 H2O(l) + 2 e - ¡ 2 Ni(OH)2(s) + 2 OH - (aq) Cd(s) + 2 OH - (aq) ¡ Cd(OH)2(s) + 2 e -
As in the lead-acid battery, the solid reaction products adhere to the electrodes, which permits the electrode reactions to be reversed during charging. A single nicad voltaic cell has a voltage of 1.30 V. Nicad battery packs typically contain three or more cells in series to produce the higher voltages needed by most electronic devices. Two drawbacks to nickel-cadmium batteries are that cadmium is both toxic and dense. Its use increases battery weight and provides an environmental hazard—roughly 1.5 billion nickel-cadmium batteries are produced annually, and these must eventually be recycled as they lose their ability to be recharged. Some of the problem has been alleviated by the development of the nickel-metal-hydride (NiMH) battery. The cathode reaction is the same as that for nickel-cadmium batteries, but the anode reaction is very different. The anode consists of a metal alloy, such as ZrNi2, that has the ability to absorb hydrogen atoms. During oxidation at the anode, the hydrogen atoms lose electrons, and the resultant H + ions react with OH - ions to form H2O, a process that is reversed during charging. The current generation of hybrid gas–electric automobiles use NiMH batteries, which are recharged by the electric motor while braking. These batteries can last up to 8 years. The newest rechargeable battery to receive large use in consumer devices is the lithium-ion (Li-ion) battery. You will find this battery in cell phones and laptop computers. Because lithium is a very light element, Li-ion batteries achieve a greater energy density—the amount of energy stored per unit mass—than nickel-based batteries. The technology of Li-ion batteries is based on the ability of Li + ions to be inserted into and removed from certain layered solids. For example, Li + ions can be inserted reversibly into layers of graphite (Figure 12.30). In most commercial cells, one electrode is graphite or some other carbon-based material, and the other is usually made of lithium cobalt oxide (LiCoO2). When the cell is charged, cobalt ions are oxidized and Li + ions migrate into the graphite. During discharge, when the battery is producing electricity for use, the Li + ions spontaneously migrate from the graphite anode to the cathode, enabling electrons to flow through the external circuit. An Li-ion battery produces a maximum voltage of 3.7 V, considerably higher than typical 1.5-V alkaline batteries.
Hydrogen Fuel Cells The thermal energy released by burning fuels can be converted to electrical energy. The thermal energy may convert water to steam, for instance, which drives a turbine that in turn drives an electrical generator. Typically, a maximum of only 40% of the energy from combustion is converted to electricity in this manner; the remainder is lost as heat. The direct production of electricity from fuels by a voltaic cell could, in principle, yield a higher rate of conversion of chemical energy to electrical energy. Voltaic cells that perform this conversion using conventional fuels, such as H2 and CH4, are called fuel cells. Strictly speaking, fuel cells are not batteries because they are not self-contained systems—the fuel must be continuously supplied to generate electricity.
SECTION 20.8
The most common fuel-cell systems involve the reaction of H2(g) and O2(g) to form H2O(l). These cells can generate electricity twice as efficiently as the best internal combustion engine. Under acidic conditions, the reactions are Cathode: Anode: Overall:
O2(g) + 4 H + + 4 e - ¡ 2 H2O(l) 2 H2(g) ¡ 4 H + + 4 e -
857
Corrosion
Voltmeter e
H2 out
O2, H2O out
2 H2(g) + O2(g) ¡ 2 H2O(l)
These cells employ hydrogen gas as the fuel and oxygen gas from air as the oxidant and generate about 1 V. Fuel cells are often named for either the fuel or the electrolyte used. In the hydrogen-PEM fuel cell (the acronym PEM stands for either proton-exchange membrane or polymer-electrolyte membrane), the anode and cathode are separated by a membrane, which is permeable to protons but not to electrons (씰 FIGURE 20.21). The membrane therefore acts as a salt bridge. The electrodes are typically made from graphite. The hydrogen-PEM cell operates at around 80 °C. At this temperature the electrochemical reactions would normally occur very slowly, and so a thin layer of platinum on each electrode catalyzes the reactions. Many fuel cells require much higher temperatures to operate. In order to power a vehicle, multiple cells must be assembled into a fuel cell stack. The amount of power generated by a stack depends on the number and size of the fuel cells in the stack and on the surface area of the PEM. Much fuel cell research today is directed toward improving electrolytes and catalysts and toward developing cells that use fuels such as hydrocarbons and alcohols, which are not as difficult to handle and distribute as hydrogen gas.
Hⴙ H2
Hⴙ
O2
Hⴙ H2 in
O2 in Anode
Cathode PEM porous membrane
쑿 FIGURE 20.21 A hydrogen-PEM fuel cell. The proton-exchange membrane (PEM) allows H + ions generated by H2 oxidation at the anode to migrate to the cathode, where H2O is formed.
CHEMISTRY PUT TO WORK Direct Methanol Fuel Cells Even though the hydrogen fuel cell is highly developed, liquid methanol, CH3OH, is a far more attractive fuel to store and transport than hydrogen gas. Furthermore, methanol is a clean-burning liquid, and its use would require only minor modifications to existing engines and to fuel-delivery infrastructure. One of the intriguing aspects of methanol fuel is that manufacturing it could consume carbon dioxide, a source of global warming. •(Section 18.2) Methanol can be made by combining CO2 and H2, although the process is presently costly. Imagine, though, that the synthesis can be improved and that the CO2 used in the synthesis is captured from exhaust gases from power plants or even directly from the atmosphere. In such cases, the CO2 released by subsequently burning the methanol would be canceled by the carbon dioxide captured to make it. Thus, the process would be
carbon neutral, meaning that it would not increase the concentration of CO2 in the atmosphere. The prospect of a liquid fuel that could replace conventional fuels without contributing to the greenhouse effect has spurred considerable research to reduce the cost of methanol synthesis and to develop and improve methanol fuel cell technology. A direct methanol fuel cell has been developed that is similar to the hydrogen-PEM fuel cell. The reactions in the cell are Cathode:
3 2
O2(g) + 6 H + + 6 e - ¡ 3 H2O(g)
Anode:
CH3OH(l) + H2O(g) ¡ CO2(g) + 6 H + + 6 e -
Overall:
CH3OH(g) +
3 2
O2 (g) ¡ CO2(g) + 2 H2O(g)
The direct methanol fuel cell is currently too expensive and too inefficient to be used in passenger cars. Nevertheless, small methanol fuel cells could appear in mobile devices such as computers or cell phones in the near future.
20.8 | CORROSION In this section we examine the undesirable redox reactions that lead to corrosion of metals. Corrosion reactions are spontaneous redox reactions in which a metal is attacked by some substance in its environment and converted to an unwanted compound. For nearly all metals, oxidation is thermodynamically favorable in air at room temperature. When oxidation of a metal object is not inhibited, it can destroy the object. Oxidation can form an insulating protective oxide layer, however, that prevents further reaction of the underlying metal. Based on the standard reduction potential for Al3 + , for example, we expect aluminum metal to be readily oxidized. The many aluminum soft-drink and beer cans that litter the environment are ample evidence, however, that
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aluminum undergoes only very slow chemical corrosion. The exceptional stability of this active metal in air is due to the formation of a thin protective coat of oxide—a hydrated form of Al2O3—on the metal surface. The oxide coat is impermeable to O2 or H2O and so protects the underlying metal from further corrosion. Magnesium metal is similarly protected, and some metal alloys, such as stainless steel, likewise form protective impervious oxide coats.
Corrosion of Iron (Rusting) The rusting of iron is a familiar corrosion process that carries a significant economic impact. Up to 20% of the iron produced annually in the United States is used to replace iron objects that have been discarded because of rust damage. Rusting of iron requires both oxygen and water, and the process can be accelerated by other factors such as pH, presence of salts, contact with metals more difficult to oxidize than iron, and stress on the iron. The corrosion process involves oxidation and reduction, and the metal conducts electricity. Thus, electrons can move through the metal from a region where oxidation occurs to a region where reduction occurs, as in voltaic cells. Because the standard reduction potential for reduction of Fe 2 + (aq) is less positive than that for reduction of O2, Fe(s) can be oxidized by O2(g): Cathode:
O2(g) + 4 H + (aq) + 4 e - ¡ 2 H2O(l)
° = 1.23 V E red
Fe(s) ¡ Fe 2 + (aq) + 2 e -
Anode:
° = -0.44 V E red
A portion of the iron, often associated with a dent or region of strain, can serve as an anode at which Fe is oxidized to Fe 2 + (쑼 FIGURE 20.22). The electrons produced in the oxidation migrate through the metal from this anodic region to another portion of the surface, which serves as the cathode where O2 is reduced. The reduction of O2 requires H + , so lowering the concentration of H + (increasing the pH) makes O2 reduction less favorable. Iron in contact with a solution whose pH is greater than 9 does not corrode. The Fe 2 + formed at the anode is eventually oxidized to Fe 3 + , which forms the hydrated iron(III) oxide known as rust:* 4 Fe 2 + (aq) + O2(g) + 4 H2O(l) + 2 xH2O(l) ¡ 2 Fe 2O3 # xH2O(s) + 8 H + (aq)
GO FIGURE
What is the oxidizing agent in this corrosion reaction?
1 Fe oxidized at anode region of metal
Water drop
4
Fe2 oxidized to Fe3, rust (Fe2O3) forms
O2
Fe2(aq)
Fe
씰 FIGURE 20.22 Corrosion of iron in contact with water. One region of the iron acts as the cathode and another region acts as the anode.
Fe2 2 e
2 Electrons from Fe oxidation migrate to region acting as cathode
3
O2 reduced at cathode region
e O2 4 H 4 e or O2 2 H2O 4 e
2 H2O 4 OH
*Frequently, metal compounds obtained from aqueous solution have water associated with them. For example, copper(II) sulfate crystallizes from water with 5 mol of water per mole of CuSO4. We represent this substance by the formula CuSO4 • 5H2O. Such compounds are called hydrates. • (Section 13.1) Rust is a hydrate of iron(III) oxide with a variable amount of water of hydration. We represent this variable water content by writing the formula Fe 2O3 • xH2O.
SECTION 20.8
859
Water drop
Zinc coating is oxidized (anode)
Zn2(aq)
Zn
Corrosion
O2
Zn2 2 e e O2 4 H 4 e
2 H2O
Iron is not oxidized (cathode)
씱 FIGURE 20.23 Cathodic protection of iron in contact with zinc. The standard ° Fe2 + = -0.440 V, reduction potentials are Ered, ° Zn2 + = -0.763 V, making the zinc more Ered, readily oxidized.
Because the cathode is generally the area having the largest supply of O2, rust often deposits there. If you look closely at a shovel after it has stood outside in the moist air with wet dirt adhered to its blade, you may notice that pitting has occurred under the dirt but that rust has formed elsewhere, where O2 is more readily available. The enhanced corrosion caused by the presence of salts is usually evident on autos in areas where roads are heavily salted during winter. Like a salt bridge in a voltaic cell, the ions of the salt provide the electrolyte necessary to complete the electrical circuit.
Preventing Corrosion of Iron Objects made of iron are often covered with a coat of paint or another metal to protect against corrosion. Covering the surface with paint prevents oxygen and water from reaching the iron surface. If the coating is broken, however, and the iron exposed to oxygen and water, corrosion begins as the iron is oxidized. With galvanized iron, which is iron coated with a thin layer of Ground Soil zinc, the iron is protected from corrosion even after the surface level electrolyte coat is broken. The standard reduction potentials are Fe 2 + (aq) + 2 e - ¡ Fe(s)
Iron pipe, cathode
° = -0.44 V E red
° = -0.76 V Zn2 + (aq) + 2 e - ¡ Zn(s) E red ° for Fe 2 + is less negative (more positive) than E red ° for Because E red 2+ Zn , Zn(s) is more readily oxidized than Fe(s). Thus, even if the zinc coating is broken and the galvanized iron is exposed to oxygen and water, as in 쑿 FIGURE 20.23, the zinc serves as the anode and is corroded (oxidized) instead of the iron. The iron serves as the cathode at which O2 is reduced. Protecting a metal from corrosion by making it the cathode in an electrochemical cell is known as cathodic protection. The metal that is oxidized while protecting the cathode is called the sacrificial anode. Underground pipelines and storage tanks made of iron are often protected against corrosion by making the iron the cathode of a voltaic cell. For example, pieces of a metal that is more easily oxidized than iron, such as magnesium ° = -2.37 V), are buried near the pipe or storage tank and (E red connected to it by wire (씰 FIGURE 20.24). In moist soil, where corrosion can occur, the sacrificial metal serves as the anode, and the pipe or tank experiences cathodic protection.
Insulated copper wire 30 cm
Soldered connection
Sacrificial magnesium anode 쑿 FIGURE 20.24 Cathodic protection of an iron pipe. A mixture of gypsum, sodium sulfate, and clay surrounds the sacrificial magnesium anode to promote conductivity of ions.
GIVE IT SOME THOUGHT Based on the values in Table 20.1, which of these metals could provide cathodic protection to iron: Al, Cu, Ni, Zn?
860
CHAPTER 20
e
Electrochemistry
e
20.9 | ELECTROLYSIS
Voltage source Anode
Cathode
Cl
Na
Cl2(g)
Na(l)
Molten NaCl 2 Cl
Cl2(g) 2 e 2 Na 2 e
2 Na(l)
쑿 FIGURE 20.25 Electrolysis of molten sodium chloride. Pure NaCl melts at 801 °C.
Voltaic cells are based on spontaneous redox reactions. It is also possible for nonspontaneous redox reactions to occur, however, by using electrical energy to drive them. For example, electricity can be used to decompose molten sodium chloride into its component elements Na and Cl2. Such processes driven by an outside source of electrical energy are called electrolysis reactions and take place in electrolytic cells. An electrolytic cell consists of two electrodes immersed either in a molten salt or in a solution. A battery or some other source of electrical energy acts as an electron pump, pushing electrons into one electrode and pulling them from the other. Just as in voltaic cells, the electrode at which reduction occurs is called the cathode, and the electrode at which oxidation occurs is called the anode. In the electrolysis of molten NaCl, Na + ions pick up electrons and are reduced to Na at the cathode 씱 FIGURE 20.25. As Na + ions near the cathode are depleted, additional Na + ions migrate in. Similarly, there is net movement of Cl - ions to the anode where they are oxidized. The electrode reactions for the electrolysis are 2 Na + (l) + 2 e - ¡ 2 Na(l) 2 Cl - (l) ¡ Cl 2(g) + 2 e -
Cathode: Anode:
2 Na + (l) + 2 Cl - (l) ¡ 2 Na(l) + Cl 2(g)
e
Voltage source
Nickel anode
Ni2 Ni2
Ni(s)
Anode Ni2(aq) 2 e
Notice how the energy source is connected to the electrodes in Figure 20.25. The positive terminal is connected to the anode and the negative terminal is connected to the cathode, which forces electrons to move from the anode to the cathode. Because of the high melting points of ionic substances, the electrolysis of molten salts requires very high temperatures. Do we obtain the same products if we electrolyze the aqueous solution of a salt instead of the molten salt? Frequently the answer is no because water itself might be oxidized to form O2 or reduced to form H2 rather than the ions of the salt. In our examples of the electrolysis of NaCl, the electrodes are inert; they do not react but merely serve as the surface where oxidation and reduction occur. Several practical applications of electrochemistry, however, are based on active electrodes—electrodes that participate in the electrolysis process. Electroplating, for example, uses electrolysis to deposit a thin layer of one metal on another metal to improve beauty or resistance to corrosion. Examples include electroplating nickel or chromium onto steel and electroplating a precious metal like silver onto a less expensive one. 씱 FIGURE 20.26 illustrates an electrolytic cell for electroplating nickel onto a piece of steel. The anode is a strip of nickel metal, and the cathode is the steel. The electrodes are immersed in a solution of NiSO4(aq). When an external voltage is applied, reduction occurs at the cathode. The standard reduction e ° = -0.28 V) is less negative than that of potential of Ni2 + (E red Steel ° = -0.83 V), so Ni2 + is preferentially reduced, depositH2O (E red object ing a layer of nickel metal on the steel cathode. At the anode, the nickel metal is oxidized. To explain this behavior, we need to compare the substances in contact with the anode, H2O and NiSO4(aq), with the anode material, Ni. For the NiSO4(aq) solution, neither Ni2 + nor SO42- can be oxidized (because both already have their elements in their highest possible oxidation state). The H2O solvent and the Ni atoms in the anode, however, can both undergo oxidation:
Cathode Ni2(aq) 2 e
Ni(s)
쑿 FIGURE 20.26 Electrolytic cell with an active metal electrode. Nickel dissolves from the anode to form Ni2 + (aq). At the cathode Ni2 + (aq) is reduced and forms a nickel “plate” on the steel cathode.
2 H2O(l) ¡ O2(g) + 4 H + (aq) + 4 e Ni(s) ¡ Ni2 + (aq) + 2 e -
° = +1.23 V E red ° = -0.28 V E red
We saw in Section 20.4 that the half-reaction with the more ° undergoes oxidation more readily. (Remember negative E red Figure 20.12: The strongest reducing agents, which are the sub° values.) stances oxidized most readily, have the most negative E red
SECTION 20.9
° = -0.28 V, that is oxidized at the anode rather than Thus, it is the Ni(s), with its E red the H2O. If we look at the overall reaction, it appears as if nothing has been accomplished. However, this is not true because Ni atoms are transferred from the Ni anode to the steel cathode, plating the steel with a thin layer of nickel atoms. The standard emf for the overall reaction is
Electrolysis
861
Current (amperes) and time (seconds)
° = E red ° (cathode) - E red ° (anode) = (-0.28 V) - (-0.28 V) = 0 E cell
Quantity of charge (coulombs)
Because the standard emf is zero, only a small emf is needed to cause the transfer of nickel atoms from one electrode to the other.
Quantitative Aspects of Electrolysis Moles of electrons
The stoichiometry of a half-reaction shows how many electrons are needed to achieve an electrolytic process. For example, the reduction of Na + to Na is a one-electron process: Na + + e - ¡ Na Thus, 1 mol of electrons plates out 1 mol of Na metal, 2 mol of electrons plate out 2 mol of Na metal, and so forth. Similarly, 2 mol of electrons are required to produce 1 mol of Cu from Cu2 + , and 3 mol of electrons are required to produce 1 mol of Al from Al3 + :
Moles of substance oxidized or reduced
Cu2 + + 2 e - ¡ Cu Al3 + + 3 e - ¡ Al For any half-reaction, the amount of substance reduced or oxidized in an electrolytic cell is directly proportional to the number of electrons passed into the cell. The quantity of charge passing through an electrical circuit, such as that in an electrolytic cell, is generally measured in coulombs. As noted in Section 20.5, the charge on 1 mol of electrons is 96,485 C. A coulomb is the quantity of charge passing a point in a circuit in 1 s when the current is 1 ampere (A). Therefore, the number of coulombs passing through a cell can be obtained by multiplying the current in amperes by the elapsed time in seconds. Coulombs = amperes * seconds
[20.21]
Grams of substance oxidized or reduced 쑿 FIGURE 20.27 Relationship between charge and amount of reactant and product in electrolysis reactions.
씰 FIGURE 20.27 shows how the quantities of substances produced or consumed in electrolysis are related to the quantity of electrical charge used. The same relationships can also be applied to voltaic cells. In other words, electrons can be thought of as “reagents” in electrolysis reactions.
SAMPLE EXERCISE 20.14
Relating Electrical Charge and Quantity of Electrolysis
Calculate the number of grams of aluminum produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A. SOLUTION Analyze We are told that AlCl3 is electrolyzed to form Al and asked to calculate the number of grams of Al produced in 1.00 h with 10.0 A. Plan Figure 20.27 provides a road map for this problem. First, the product of the amperage and the time in seconds gives the number of coulombs of electrical charge being used (Equation 20.21). Second, the coulombs can be converted with Faraday’s constant (F = 96,485 C>mol electrons) to tell us the number of moles of
electrons being supplied. Third, reduction of 1 mol of Al3 + to Al requires 3 mol of electrons. Hence, we can use the number of moles of electrons to calculate the number of moles of Al metal it produces. Finally, we convert moles of Al into grams.
Solve First, we calculate the coulombs of electrical charge passed into the electrolytic cell:
Coulombs = amperes * seconds = (10.0 A)(1.00 h) a
Second, we calculate the number of moles of electrons that pass into the cell:
Moles e - = (3.60 * 104 C)a
Third, we relate number of moles of electrons to number of moles of aluminum formed, using the half-reaction for the reduction of Al3+ :
Al3 + + 3 e - ¡ Al
3600 s b = 3.60 * 104 C h
1 mol e b = 0.373 mol e 96,485 C
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Thus, 3 mol of electrons are required to form 1 mol of Al: Finally, we convert moles to grams: Because each step involves multiplication by a new factor, we can combine all the steps:
1 mol Al b = 0.124 mol Al 3 mol e 27.0 g Al b = 3.36 g Al Grams Al = (0.124 mol Al) a 1 mol Al
Moles Al = (0.373 mol e - )a
Grams Al = (3.60 * 104 C)a
PRACTICE EXERCISE (a) The half-reaction for formation of magnesium metal upon electrolysis of molten MgCl2 is Mg 2 + + 2 e - ¡ Mg. Calculate the mass of magnesium formed upon passage of a current of 60.0 A for a
27.0 g Al 1 mole e 1 mol Al ba b = 3.36 g Al -ba 96,485 C 3 mol e 1 mol Al
period of 4.00 * 103 s. (b) How many seconds would be required to produce 50.0 g of Mg from MgCl2 if the current is 100.0 A?
Answers: (a) 30.2 g of Mg, (b) 3.97 * 103 s
CHEMISTRY PUT TO WORK Electrometallurgy of Aluminum Many processes used to produce or refine metals are based on electrolysis. Collectively these processes are referred to as electrometallurgy. Electrometallurgical procedures can be broadly differentiated according to whether they involve electrolysis of a molten salt or of an aqueous solution. Electrolytic methods using molten salts are important for obtaining the more active metals, such as sodium, magnesium, and aluminum. These metals cannot be obtained from aqueous solution because water is more easily reduced than the metal ions. The standard reduction potentials of water under both acidic ° = 0.00 V) and basic (E red ° = -0.83 V) conditions are more (E red ° = -2.71 V), Mg 2+ (E red ° = positive than those of Na + (E red 3+ ° = -1.66 V). -2.37 V), and Al (E red Historically, obtaining aluminum metal has been a challenge. It is obtained from bauxite ore, which is chemically treated to concentrate aluminum oxide (Al2O3). The melting point of aluminum oxide is above 2000 ° C, which is too high to permit its use as a molten medium for electrolysis. The electrolytic process used commercially to produce aluminum is the Hall-Héroult process, named after its inventors, Charles M. Hall and Paul Héroult. Hall (쑼 FIGURE 20.28) began working on the problem of reducing aluminum in about 1885 after he had learned from a professor of the difficulty of reducing ores of very active metals. Before the development of an electrolytic process, aluminum was obtained by a chemical reduction using sodium or potassium as the reducing agent, a costly procedure that made aluminum metal expensive. As late as 1852, the cost of aluminum was $545 per pound, far greater than the cost of gold. During the Paris Exposition in 1855, aluminum was exhibited as a rare metal, even though it is the third most abundant element in Earth’s crust. 씱 FIGURE 20.28 Charles M. Hall (1863–1914) as a young man.
Graphite anodes
Al2O3 dissolved in molten cryolite Molten aluminum metal
Carbon-lined iron anode
씱 FIGURE 20.29 The Hall-Héroult process. Because molten aluminum is denser than the mixture of cryolite (Na3AlF6) and Al2O 3, the metal collects at the bottom of the cell.
Hall, who was 21 years old when he began his research, utilized handmade and borrowed equipment in his studies and used a woodshed near his Ohio home as his laboratory. In about a year’s time he developed an electrolytic procedure using an ionic compound that melts to form a conducting medium that dissolves Al2O3 but does not interfere with the electrolysis reactions. The ionic compound he selected was the relatively rare mineral cryolite (Na3AlF6). Héroult, who was the same age as Hall, independently made the same discovery in France at about the same time. Because of the research of these two unknown young scientists, large-scale production of aluminum became commercially feasible, and aluminum became a common and familiar metal. Indeed, the factory that Hall subsequently built to produce aluminum evolved into Alcoa Corporation. In the Hall-Héroult process, Al2O3 is dissolved in molten cryolite, which melts at 1012 ° C and is an effective electrical conductor (쑿 FIGURE 20.29). Graphite rods are employed as anodes and are consumed in the electrolysis: Anode: Cathode:
C(s) + 2 O2-(l) ¡ CO2(g) + 4 e 3 e - + Al3 + (l) ¡ Al(l)
A large amount of electrical energy is needed in the Hall-Héroult process with the result that the aluminum industry consumes about 2% of the electrical energy generated in the United States. Because recycled aluminum requires only 5% of the energy needed to produce “new” aluminum, considerable energy savings can be realized by increasing the amount of aluminum recycled. Approximately 42% of postconsumer aluminum is recycled in the United States.
SECTION 20.9
SAMPLE INTEGRATIVE EXERCISE
Putting Concepts Together
The Ksp at 298 K for iron(II) fluoride is 2.4 * 10 - 6. (a) Write a half-reaction that gives the likely products of the two-electron reduction of FeF2(s) in water. (b) Use the Ksp value and the standard reduction potential of Fe 2 + (aq) to calculate the standard reduction potential for the half-reaction in part (a). (c) Rationalize the difference between the reduction potential in part (a) and the reduction potential for Fe 2 + (aq). SOLUTION Analyze We are going to have to combine what we know about equilibrium constants and electrochemistry to obtain reduction potentials. Plan For (a) we need to determine which ion, Fe 2 + or F - , is more likely to be reduced by two electrons and complete the overall reaction FeF2 + 2 e - ¡ ?. For (b) we need to write the chemical equation associated with the Ksp and see how it relates to E° for the reduction halfreaction in (a). For (c) we need to compare E° from (b) with the value for the reduction of Fe 2 + . Solve (a) Iron(II) fluoride is an ionic substance that consists of Fe 2 + and F - ions. We are asked to predict where two electrons could be added to FeF2. We cannot envision adding the electrons to the F - ions to form F 2 - , so it seems likely that we could reduce the Fe 2 + ions to Fe(s). We therefore predict the half-reaction FeF2(s) + 2 e - ¡ Fe(s) + 2 F - (aq) (b) The Ksp for FeF2 refers to the following equilibrium: (Section 17.4) FeF2(s) Δ Fe 2 + (aq) + 2 F - (aq) Ksp = [Fe 2 + ][F - ]2 = 2.4 * 10 - 6 We were also asked to use the standard reduction potential of Fe 2 + , whose half-reaction and standard reduction potentials are listed in Appendix E: Fe 2 + (aq) + 2 e - ¡ Fe(s) E = -0.440 V According to Hess’s law, if we can add chemical equations to get a desired equation, then we can add their associated thermodynamic state functions, like ¢H or ¢G, to determine that thermodynamic quantity for the desired reaction. • (Section 5.6) So we need to consider whether the three equations we are working with can be combined in a similar fashion. Notice that if we add the Ksp reaction to the standard reduction half-reaction for Fe 2 + , we get the halfreaction we want: FeF2(s) ¡ Fe 2 + (aq) + 2 F - (aq)
1.
2. Fe 2 + (aq) + 2 e - ¡ Fe(s) Overall:
3.
FeF2(s) + 2 e - ¡ Fe(s) + 2 F - (aq)
Reaction 3 is still a half-reaction, so we do see the free electrons. If we knew ¢G° for reactions 1 and 2, we could add them to get ¢G° for reaction 3. We can relate ¢G° to E° by ¢G° = -nFE° (Equation 20.12) and to K by ¢G° = -RT ln K (Equation 19.20, see also Figure 20.14). Furthermore, we know that K for reaction 1 is the Ksp of FeF2, and we know E° for reaction 2. Therefore we can calculate ¢G° for reactions 1 and 2: Reaction 1: ¢G° = -RT ln K = -(8.314 J>K mol)(298 K) ln(2.4 * 10 - 6) = 3.21 * 104 J>mol Reaction 2: ¢G° = -nFE° = -(2)(96,485 C>mol)(-0.440 J>C) = 8.49 * 104 J>mol (Recall that 1 volt is 1 joule per coulomb.) Then ¢G° for reaction 3, the one we want, is the sum of the ¢G° values for reactions 1 and 2: 3.21 * 104 J>mol + 8.49 * 104 J>mol = 1.17 * 105 J>mol We can convert this to E° from the relationship ¢G° = -nFE°: 1.17 * 105 J>mol = - (2)(96,485 C>mol) E° 1.17 * 105 J>mol = -0.606 J>C = -0.606 V -(2)(96,485 C>mol) (c) The standard reduction potential for FeF2 (-0.606 V) is more negative than that for Fe 2 + (-0.440 V), telling us that the reduction of FeF2 is the less favorable process. When FeF2 is reduced, we not only reduce the Fe 2 + ions but also break up the ionic solid. Because this additional energy must be overcome, the reduction of FeF2 is less favorable than the reduction of Fe 2 + . E° =
Electrolysis
863
864
CHAPTER 20
Electrochemistry
CHAPTER SUMMARY AND KEY TERMS INTRODUCTION AND SECTION 20.1 In this chapter we have focused on electrochemistry, the branch of chemistry that relates electricity and chemical reactions. Electrochemistry involves oxidation-reduction reactions, also called redox reactions. These reactions involve a change in the oxidation state of one or more elements. In every oxidation-reduction reaction one substance is oxidized (its oxidation state, or number, increases) and one substance is reduced (its oxidation state, or number, decreases). The substance that is oxidized is referred to as a reducing agent, or reductant, because it causes the reduction of some other substance. Similarly, the substance that is reduced is referred to as an oxidizing agent, or oxidant, because it causes the oxidation of some other substance.
An oxidization-reduction reaction can be balanced by dividing the reaction into two half-reactions, one for oxidation and one for reduction. A half-reaction is a balanced chemical equation that includes electrons. In oxidation half-reactions the electrons are on the product (right) side of the equation; we can envision that these electrons are transferred from a substance when it is oxidized. In reduction halfreactions the electrons are on the reactant (left) side of the equation. Each half-reaction is balanced separately, and the two are brought together with proper coefficients to balance the electrons on each side of the equation, so the electrons cancel when the half-reactions are added. SECTION 20.2
A voltaic (or galvanic) cell uses a spontaneous oxidation-reduction reaction to generate electricity. In a voltaic cell the oxidation and reduction half-reactions often occur in separate halfcells. Each half-cell has a solid surface called an electrode, where the half-reaction occurs. The electrode where oxidation occurs is called the anode; reduction occurs at the cathode. The electrons released at the anode flow through an external circuit (where they do electrical work) to the cathode. Electrical neutrality in the solution is maintained by the migration of ions between the two half-cells through a device such as a salt bridge.
SECTION 20.3
A voltaic cell generates an electromotive force (emf) that moves the electrons from the anode to the cathode through the external circuit. The origin of emf is a difference in the electrical potential energy of the two electrodes in the cell. The emf of a cell is called its cell potential, Ecell, and is measured in volts (1 V = 1 J>C). The cell potential under standard conditions is called the standard ° . emf, or the standard cell potential, and is denoted E cell ° , can be assigned for an indiA standard reduction potential, E red vidual half-reaction. This is achieved by comparing the potential of the half-reaction to that of the standard hydrogen electrode (SHE), ° = 0 V and is based on the following which is defined to have E red half-reaction: SECTION 20.4
2 H + (aq, 1 M) + 2 e - ¡ H2(g, 1 atm)
° = 0V E red
The standard cell potential of a voltaic cell is the difference between the standard reduction potentials of the half-reactions that occur at the ° = E red ° (cathode) - E red ° (anode). The cathode and the anode: E cell ° is positive for a voltaic cell. value of E cell ° is a measure of the tendency of For a reduction half-reaction, E red ° , the greater the reduction to occur; the more positive the value for E red ° provides a the tendency of the substance to be reduced. Thus, E red measure of the oxidizing strength of a substance. Substances that are strong oxidizing agents produce products that are weak reducing agents, and vice versa. The emf, E, is related to the change in the Gibbs free energy, ¢G : ¢G = -nFE, where n is the number of electrons
SECTION 20.5
transferred during the redox process and F is Faraday’s constant, defined as the quantity of electrical charge on one mole of electrons: F = 96,485 C> mol. Because E is related to ¢G, the sign of E indicates whether a redox process is spontaneous: E 7 0 indicates a spontaneous process, and E 6 0 indicates a nonspontaneous one. Because ¢G is also related to the equilibrium constant for a reaction (¢G = -RT ln K), we can relate E to K as well. The maximum amount of electrical work produced by a voltaic cell is given by the product of the total charge delivered, nF, and the emf, E: wmax = -nFE. The watt is a unit of power: 1 W = 1 J>s. Electrical work is often measured in kilowatt-hours. The emf of a redox reaction varies with temperature and with the concentrations of reactants and products. The Nernst equation relates the emf under nonstandard conditions to the standard emf and the reaction quotient Q: SECTION 20.6
E = E° - (RT>nF) ln Q = E° - (0.0592>n) log Q The factor 0.0592 is valid when T = 298K. A concentration cell is a voltaic cell in which the same half-reaction occurs at both the anode and cathode but with different concentrations of reactants in each half-cell. At equilibrium, Q = K and E = 0. SECTION 20.7 A battery is a self-contained electrochemical power source that contains one or more voltaic cells. Batteries are based on a variety of different redox reactions. Several common batteries were discussed. The lead-acid battery, the nickel-cadmium battery, the nickel-metal-hydride battery, and the lithium-ion battery are examples of rechargeable batteries. The common alkaline dry cell is not rechargeable. Fuel cells are voltaic cells that utilize redox reactions in which reactants such as H2 have to be continuously supplied to the cell to generate voltage. SECTION 20.8
Electrochemical principles help us understand
corrosion, undesirable redox reactions in which a metal is attacked by
some substance in its environment. The corrosion of iron into rust is caused by the presence of water and oxygen, and it is accelerated by the presence of electrolytes, such as road salt. The protection of a metal by putting it in contact with another metal that more readily undergoes oxidation is called cathodic protection. Galvanized iron, for example, is coated with a thin layer of zinc; because zinc is oxidized more readily than iron, the zinc serves as a sacrificial anode in the redox reaction. An electrolysis reaction, which is carried out in an electrolytic cell, employs an external source of electricity to drive a SECTION 20.9
nonspontaneous electrochemical reaction. The current-carrying medium within an electrolytic cell may be either a molten salt or an electrolyte solution. The products of electrolysis can generally be predicted by comparing the reduction potentials associated with possible oxidation and reduction processes. The electrodes in an electrolytic cell can be active, meaning that the electrode can be involved in the electrolysis reaction. Active electrodes are important in electroplating and in metallurgical processes. The quantity of substances formed during electrolysis can be calculated by considering the number of electrons involved in the redox reaction and the amount of electrical charge that passes into the cell. The amount of electrical charge is measured in coulombs and is related to the magnitude of the current and the time it flows (1 C = 1 A-s). Electrometallurgy is the use of electrolytic methods to prepare or purify a metallic element. Aluminum is obtained in the Hall-Héroult process by electrolysis of Al2O3 in molten cryolite (Na3AlF6).
Exercises
865
KEY SKILLS • Identify oxidation, reduction, oxidizing agent, and reducing agent in a chemical equation. (Section 20.1) • Complete and balance redox equations using the method of half-reactions. (Section 20.2) • Sketch a voltaic cell and identify its cathode, anode, and the directions that electrons and ions move. (Section 20.3) ° , from standard reduction potentials. (Section 20.4) • Calculate standard emfs (cell potentials), E cell
• Use reduction potentials to predict whether a redox reaction is spontaneous. (Section 20.4) ° to ¢G° and equilibrium constants. (Section 20.5) • Relate E cell
• Calculate emf under nonstandard conditions. (Section 20.6) • Describe the components of common batteries and fuel cells. (Section 20.7) • Explain how corrosion occurs and how it is prevented by cathodic protection. (Section 20.8) • Describe the reactions in electrolytic cells. (Section 20.9) • Relate amounts of products and reactants in redox reactions to electrical charge. (Section 20.9)
KEY EQUATIONS ° = E red ° (cathode) - E red ° (anode) • E cell
[20.8]
• ¢G = -nFE 0.0592 V log Q (at 298 K) • E = E° n
[20.11]
Relating standard emf to standard reduction potentials of the reduction (cathode) and oxidation (anode) half-reactions Relating free-energy change and emf
[20.18]
The Nernst equation, expressing the effect of concentration on cell potential
EXERCISES VISUALIZING CONCEPTS 20.1 In the Brønsted-Lowry concept of acids and bases, acid–base reactions are viewed as proton-transfer reactions. The stronger the acid, the weaker is its conjugate base. In what ways are redox reactions analogous? [Sections 20.1 and 20.2] 20.2 You may have heard that “antioxidants” are good for your health. Based on what you have learned in this chapter, what do you deduce an “antioxidant” is? [Sections 20.1 and 20.2]
You begin with the incomplete cell pictured here in which the electrodes are immersed in water.
A
Voltmeter
B
20.3 The diagram that follows represents a molecular view of a process occurring at an electrode in a voltaic cell.
(a) Does the process represent oxidation or reduction? (b) Is the electrode the anode or cathode? (c) Why are the atoms in the electrode represented by larger spheres than the ions in the solution? [Section 20.3] 20.4 Assume that you want to construct a voltaic cell that uses the following half-reactions: ° = -0.10 V A2 + (aq) + 2 e - ¡ A(s) E red ° = -1.10 V B2 + (aq) + 2 e - ¡ B(s) E red
(a) What additions must you make to the cell for it to generate a standard emf? (b) Which electrode functions as the cathode? (c) Which direction do electrons move through the external circuit? (d) What voltage will the cell generate under standard conditions? [Sections 20.3 and 20.4] 20.5 For a spontaneous reaction A(aq) + B(aq) ¡ A-(aq) + B + (aq), answer the following questions: (a) If you made a voltaic cell out of this reaction, what halfreaction would be occurring at the cathode, and what half-reaction would be occurring at the anode? (b) Which half-reaction from (a) is higher in potential energy? ° ? [Section 20.3] (c) What is the sign of E cell
866
CHAPTER 20
Electrochemistry
20.6 Consider the following table of standard electrode potentials for a series of hypothetical reactions in aqueous solution: Reduction Half-Reaction
E°(V)
A+ (aq) + e - ¡ A(s) B2 + (aq) + 2e - ¡ B(s) C3 + (aq) + e - ¡ C2 + (aq) D3 + (aq) + 3e - ¡ D(s)
1.33 0.87 -0.12 -1.59
1
20.7 Consider a redox reaction for which E° is a negative number. (a) What is the sign of ¢G° for the reaction? (b) Will the equilibrium constant for the reaction be larger or smaller than 1? (c) Can an electrochemical cell based on this reaction accomplish work on its surroundings? [Section 20.5] 20.8 Consider the following voltaic cell: Voltmeter e
Salt bridge
2
Ered
(a) Which substance is the strongest oxidizing agent? Which is weakest? (b) Which substance is the strongest reducing agent? Which is weakest? (c) Which substance(s) can oxidize C2+? [Sections 20.4 and 20.5]
Fe(s)
20.9 Consider the half-reaction Ag + (aq) + e - ¡ Ag(s). (a) Which of the lines in the following diagram indicates how the reduction potential varies with the concentration of Ag + ? (b) What is the value of Ered when log[Ag + ] = 0? [Section 20.6]
3 log[Ag] 20.10 Draw a generic picture of a fuel cell. What is the main difference between it and a battery, regardless of the redox reactions that occur inside? [Section 20.7] 20.11 How does a zinc coating on iron protect the iron from unwanted oxidation? [Section 20.8] 20.12 Magnesium is produced commercially by electrolysis from a molten salt using a cell similar to the one shown here. (a) What salt is used as the electrolyte? (b) Which electrode is the anode, and which one is the cathode? (c) Write the overall cell reaction and individual half-reactions. (d) What precautions would need to be taken with respect to the magnesium formed? [Section 20.9] Voltage source
Ag(s) Steel electrode
Cl2(g) out
Carbon electrode
1 M Fe2 1 M Ag (a) Which electrode is the cathode? (b) How would you determine the standard emf generated by this cell? (c) What is the change in the cell voltage when the ion concentrations in the cathode half-cell are increased by a factor of 10? (d) What is the change in the cell voltage when the ion concentrations in the anode half-cell are increased by a factor of 10? [Sections 20.4 and 20.6]
Liquid Mg
Electrolyte
OXIDATION-REDUCTION REACTIONS (section 20.1) 20.13 (a) What is meant by the term oxidation? (b) On which side of an oxidation half-reaction do the electrons appear? (c) What is meant by the term oxidant? (d) What is meant by the term oxidizing agent? 20.14 (a) What is meant by the term reduction? (b) On which side of a reduction half-reaction do the electrons appear? (c) What is meant by the term reductant? (d) What is meant by the term reducing agent? 20.15 Indicate whether each of the following statements is true or false: (a) If something is oxidized, it is formally losing electrons.
(b) For the reaction Fe 3 + (aq) + Co2 + (aq) ¡ Fe 2 + (aq) + Co3 + (aq), Fe 3 + (aq) is the reducing agent and Co2 + (aq) is the oxidizing agent. (c) If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a redox reaction. 20.16 Indicate whether each of the following statements is true or false: (a) If something is reduced, it is formally losing electrons. (b) A reducing agent gets oxidized as it reacts. (c) Oxidizing agents can convert CO into CO2.
Exercises 20.17 In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case. (a) I2O5(s) + 5 CO(g) ¡ I2(s) + 5 CO2(g) (b) 2 Hg 2 + (aq) + N2H4 (aq) ¡ 2 Hg(l) + N2(g) + 4 H + (aq) (c) 3 H2S(aq) + 2 H + (aq) + 2 NO3 - (aq) ¡ 3 S(s) + 2 NO(g) + 4 H2O(l) (d) Ba2 + (aq) + 2 OH - (aq) + H2O2(aq) + 2 ClO2(aq) ¡ Ba(ClO2)2 (s) + 2 H2O(l) + O2(g)
867
20.18 Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) PBr3(l) + 3 H2O(l) ¡ H3PO3(aq) + 3 HBr(aq) (b) NaI(aq) + 3 HOCl(aq) ¡ NaIO3(aq) + 3 HCl(aq) (c) 3 SO2(g) + 2 HNO3(aq) + 2 H2O(l) ¡ 3 H2SO4(aq) + 2 NO(g) (d) 2 H2SO4(aq) + 2 NaBr(s) ¡ Br2(l) + SO2(g) + Na2SO4(aq) + 2 H2O(l)
BALANCING OXIDATION-REDUCTION REACTIONS (section 20.2) 20.19 At 900 °C titanium tetrachloride vapor reacts with molten magnesium metal to form solid titanium metal and molten magnesium chloride. (a) Write a balanced equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance is the reductant, and which is the oxidant? 20.20 Hydrazine (N2H4) and dinitrogen tetroxide (N2O4) form a self-igniting mixture that has been used as a rocket propellant. The reaction products are N2 and H2O. (a) Write a balanced chemical equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance serves as the reducing agent and which as the oxidizing agent? 20.21 Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) Sn2 + (aq) ¡ Sn4 + (aq) (acidic solution) (b) TiO2(s) ¡ Ti2 + (aq) (acidic solution) (c) ClO3 - (aq) ¡ Cl - (aq) (acidic solution) (d) N2(g) ¡ NH4 + (aq) (acidic solution) (e) OH - (aq) ¡ O2(g) (basic solution) (f SO32 - (aq) ¡ SO42 - (aq) (basic solution) (g) N2(g) ¡ NH3(g) (basic solution) 20.22 Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) Mo3 + (aq) ¡ Mo(s) (acidic solution) (b) H2SO3(aq) ¡ SO42 - (aq) (acidic solution) (c) NO3 - (aq) ¡ NO(g) (acidic solution) (d) O2(g) ¡ H2O(l) (acidic solution) (e) O2(g) ¡ H2O(l) (basic solution) (f) Mn2 + (aq) ¡ MnO2(s) (basic solution) (g) Cr(OH)3(s) ¡ CrO42 - (aq) (basic solution)
20.23 Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) Cr2O72 - (aq) + I - (aq) ¡ Cr 3 + (aq) + IO3 - (aq) (acidic solution) (b) MnO4 (aq) + CH3OH(aq) ¡ Mn2 + (aq) + HCO2H(aq) (acidic solution) (c) I2(s) + OCl - (aq) ¡ IO3 - (aq) + Cl - (aq) (acidic solution) (d) As2O3(s) + NO3 (aq) ¡ H3AsO4(aq) + N2O3(aq) (acidic solution) (e) MnO4 (aq) + Br (aq) ¡ MnO2(s) + BrO3 - (aq) (basic solution) 2(f) Pb(OH)4 (aq) + ClO (aq) ¡ PbO2(s) + Cl - (aq) (basic solution) 20.24 Complete and balance the following equations, and identify the oxidizing and reducing agents. (Recall that the O atoms in hydrogen peroxide, H2O2, have an atypical oxidation state.) (a) NO2 - (aq) + Cr2O72 - (aq) ¡ Cr 3 + (aq) + NO3 - (aq) (acidic solution) (b) S(s) + HNO3(aq) ¡ H2SO3(aq) + N2O(g) (acidic solution) (c) Cr2O72 - (aq) + CH3OH(aq) ¡ HCO2H(aq) + Cr 3 + (aq) (acidic solution) (d) BrO3 (aq) + N2H4(g) ¡ Br - (aq) + N2(g) (acidic solution) (e) NO2 - (aq) + Al(s) ¡ NH4 + (aq) + AlO2 - (aq) (basic solution) (f) H2O2(aq) + ClO2(aq) ¡ ClO2 - (aq) + O2(g) (basic solution)
VOLTAIC CELLS (section 20.3) 20.25 (a) What are the similarities and differences between Figure 20.3 and Figure 20.4? (b) Why are Na + ions drawn into the cathode half-cell as the voltaic cell shown in Figure 20.5 operates? 20.26 (a) What is the role of the porous glass disc shown in Figure 20.4? (b) Why do NO3 - ions migrate into the anode half-cell as the voltaic cell shown in Figure 20.5 operates? 20.27 A voltaic cell similar to that shown in Figure 20.5 is constructed. One electrode half-cell consists of a silver strip placed in a
solution of AgNO3, and the other has an iron strip placed in a solution of FeCl2. The overall cell reaction is Fe(s) + 2 Ag + (aq) ¡ Fe 2 + (aq) + 2 Ag(s) (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the silver electrode to the iron electrode or from the iron to the silver? (f) In which directions do the cations and anions migrate through the solution?
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20.28 A voltaic cell similar to that shown in Figure 20.5 is constructed. One half-cell consists of an aluminum strip placed in a solution of Al(NO3)3, and the other has a nickel strip placed in a solution of NiSO4. The overall cell reaction is 2 Al(s) + 3 Ni2 + (aq) ¡ 2 Al3 + (aq) + 3 Ni(s) (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells.
(c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the nickel electrode or from the nickel to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the Al is not coated with its oxide.
CELL POTENTIALS UNDER STANDARD CONDITIONS (section 20.4) 20.29 (a) What does the term electromotive force mean? (b) What is the definition of the volt? (c) What does the term cell potential mean? 20.30 (a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons? (b) What are the units for electrical potential? How does this unit relate to energy expressed in joules? (c) What is special about a standard cell potential? 20.31 (a) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the cathode of a voltaic cell. (b) What is standard about the standard hydrogen electrode? (c) What is the role of the platinum foil in a standard hydrogen electrode? 20.32 (a) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (b) The platinum electrode in a standard hydrogen electrode is specially prepared to have a large surface area. Why is this important? (c) Sketch a standard hydrogen electrode. 20.33 (a) What is a standard reduction potential? (b) What is the standard reduction potential of a standard hydrogen electrode? 20.34 (a) Why is it impossible to measure the standard reduction potential of a single half-reaction? (b) Describe how the standard reduction potential of a half-reaction can be determined. 20.35 A voltaic cell that uses the reaction Tl3 + (aq) + 2 Cr 2 + (aq) ¡ Tl + (aq) + 2 Cr 3 + (aq) has a measured standard cell potential of + 1.19 V. (a) Write the two half-cell reactions. (b) By using data from Appendix E, ° for the reduction of Tl3 + (aq) to Tl + (aq). (c) determine E red Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow. 20.36 A voltaic cell that uses the reaction PdCl 42 - (aq) + Cd(s) ¡ Pd(s) + 4 Cl - (aq) + Cd2 + (aq) has a measured standard cell potential of +1.03 V. (a) Write the two half-cell reactions. (b) By using data from Appendix E, ° for the reaction involving Pd. (c) Sketch the determine E red voltaic cell, label the anode and cathode, and indicate the direction of electron flow. 20.37 Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions: (a) Cl 2(g) + 2 I - (aq) ¡ 2 Cl - (aq) + I2(s) (b) Ni(s) + 2 Ce 4 + (aq) ¡ Ni2 + (aq) + 2 Ce 3 + (aq) (c) Fe(s) + 2 Fe 3 + (aq) ¡ 3 Fe 2 + (aq) (d) 2 NO3 -(aq) + 8 H + (aq) + 3 Cu(s) ¡ 2 NO(g) + 4 H2O(l) + 3 Cu2 + (aq)
20.38 Using data in Appendix E, calculate the standard emf for each of the following reactions: (a) H2(g) + F2(g) ¡ 2 H + (aq) + 2 F - (aq) (b) Cu2 + (aq) + Ca(s) ¡ Cu(s) + Ca2 + (aq) (c) 3 Fe 2 + (aq) ¡ Fe(s) + 2 Fe 3 + (aq) (d) 2 ClO3 -(aq) + 10 Br - (aq) + 12 H + (aq) ¡ Cl 2(g) + 5 Br2(l) + 6 H2O(l) 20.39 The standard reduction potentials of the following halfreactions are given in Appendix E: Ag + (aq) + e - ¡ Ag(s) Cu2 + (aq) + 2 e - ¡ Cu(s) Ni2 + (aq) + 2 e - ¡ Ni(s) Cr 3 + (aq) + 3 e - ¡ Cr(s) (a) Determine which combination of these half-cell reactions leads to the cell reaction with the largest positive cell potential and calculate the value. (b) Determine which combination of these half-cell reactions leads to the cell reaction with the smallest positive cell potential and calculate the value. 20.40 Given the following half-reactions and associated standard reduction potentials: AuBr4 -(aq) + 3 e - ¡ Au(s) + 4 Br - (aq) ° = -0.858 V E red Eu3 + (aq) + e - ¡ Eu2 + (aq) ° = -0.43 V E red IO - (aq) + H2O(l) + 2 e - ¡ I -(aq) + 2 OH - (aq) ° = +0.49 V E red (a) Write the equation for the combination of these half-cell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of halfcell reactions that leads to the smallest positive emf and calculate that value. 20.41 A 1 M solution of Cu(NO3)2 is placed in a beaker with a strip of Cu metal. A 1 M solution of SnSO4 is placed in a second beaker with a strip of Sn metal. A salt bridge connects the two beakers, and wires to a voltmeter link the two metal electrodes. (a) Which electrode serves as the anode and which as the cathode? (b) Which electrode gains mass and which loses mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions? 20.42 A voltaic cell consists of a strip of cadmium metal in a solution of Cd(NO3)2 in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with Cl2 gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?
Exercises
869
STRENGTHS OF OXIDIZING AND REDUCING AGENTS (section 20.4) 20.43 From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger reducing agent: (a) Fe(s) or Mg(s) (b) Ca(s) or Al(s) (c) H2(g, acidic solution) or H2S(g) (d) BrO3 - (aq) or IO3 - (aq) 20.44 From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger oxidizing agent: (a) Cl2(g) or Br2(l) (b) Zn2 + (aq) or Cd2 + (aq) (c) Cl - (aq) or ClO3 -(aq) (d) H2O2(aq) or O3(g) 20.45 By using the data in Appendix E, determine whether each of the following substances is likely to serve as an oxidant or a reductant: (a) Cl 2(g), (b) MnO4 - (aq, acidic solution), (c) Ba(s), (d) Zn(s). 20.46 Is each of the following substances likely to serve as an oxidant or a reductant: (a) Ce 3 + (aq), (b) Ca(s), (c) ClO3 - (aq), (d) N2O5(g)?
20.47 (a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: Cr2O72 - , H2O2, Cu2 + , Cl2, O2. (b) Arrange the following in order of increasing strength as reducing agents in acidic solution: Zn, I - , Sn2 + , H2O2, Al. 20.48 Based on the data in Appendix E, (a) which of the following is the strongest oxidizing agent and which is the weakest in acidic solution:, Br2, H2O2, Zn, Cr2O72–? (b) Which of the following is the strongest reducing agent, and which is the weakest in acidic solution: F - , Zn, N2H5+ , I2, NO? 20.49 The standard reduction potential for the reduction of Eu3 + (aq) to Eu2 + (aq) is -0.43 V. Using Appendix E, which of the following substances is capable of reducing Eu3 + (aq) to Eu2 + (aq) under standard conditions: Al, Co, H2O2, N2H5 + , H2C2O4? 20.50 The standard reduction potential for the reduction of RuO4 - (aq) to RuO42 - (aq) is +0.59 V. By using Appendix E, which of the following substances can oxidize RuO42 - (aq) to RuO4 - (aq) under standard conditions: Br2(l), BrO3 - (aq), Mn2 + (aq), O2(g), Sn2 + (aq)?
FREE ENERGY AND REDOX REACTIONS (section 20.5) 20.51 Given the following reduction half-reactions: Fe
3+
(aq) + e
-
¡ Fe
2+
(aq) ° = +0.77 V E red
S2O62 - (aq) + 4 H + (aq) + 2 e - ¡ 2 H2SO3(aq) ° = +0.60 V E red N2O(g) + 2 H + (aq) + 2 e - ¡ N2(g) + H2O(l) ° = -1.77 V E red VO2 + (aq) + 2 H + (aq) + e - ¡ VO2+ + H2O(l) ° = +1.00 V E red
20.52
20.53 20.54 20.55
(a) Write balanced chemical equations for the oxidation of Fe 2 + (aq) by S2O62 - (aq), by N2O(aq), and by VO2 + (aq). (b) Calculate ¢G° for each reaction at 298 K. (c) Calculate the equilibrium constant K for each reaction at 298 K. For each of the following reactions, write a balanced equation, calculate the standard emf, calculate ¢G° at 298 K, and calculate the equilibrium constant K at 298 K. (a) Aqueous iodide ion is oxidized to I2(s) by Hg 22 + (aq). (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, Cr(OH)3(s) is oxidized to CrO42 - (aq) by ClO - (aq). If the equilibrium constant for a two-electron redox reaction at ° . 298 K is 1.5 * 10 - 4, calculate the corresponding ¢G° and E red If the equilibrium constant for a one-electron redox reaction at ° . 298 K is 8.7 * 104, calculate the corresponding ¢G° and E red Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at 298 K: (a) Fe(s) + Ni2 + (aq) ¡ Fe 2 + (aq) + Ni(s) (b) Co(s) + 2 H + (aq) ¡ Co2 + (aq) + H2(g) (c) 10 Br - (aq) + 2 MnO4 - (aq) + 16 H + (aq) ¡ 2 Mn2 + (aq) + 8 H2O(l) + 5 Br2(l)
20.56 Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at 298 K: (a) Cu(s) + 2 Ag + (aq) ¡ Cu2 + (aq) + 2 Ag(s) (b) 3 Ce 4 + (aq) + Bi(s) + H2O(l) ¡ 3 Ce 3 + (aq) + BiO + (aq) + 2 H + (aq) + (c) N2H5 (aq) + 4 Fe(CN)63 - (aq) ¡ N2(g) + 5 H + (aq) + 4 Fe(CN)64 - (aq) 20.57 A cell has a standard cell potential of +0.177 V at 298 K. What is the value of the equilibrium constant for the reaction (a) if n = 1? (b) if n = 2? (c) if n = 3? 20.58 At 298 K a cell reaction has a standard cell potential of +0.17 V. The equilibrium constant for the reaction is 5.5 * 105. What is the value of n for the reaction? 20.59 A voltaic cell is based on the reaction Sn(s) + I2(s) ¡ Sn2 + (aq) + 2 I - (aq) Under standard conditions, what is the maximum electrical work, in joules, that the cell can accomplish if 75.0 g of Sn is consumed? 20.60 Consider the voltaic cell illustrated in Figure 20.5, which is based on the cell reaction Zn(s) + Cu2 + (aq) ¡ Zn2 + (aq) + Cu(s) Under standard conditions, what is the maximum electrical work, in joules, that the cell can accomplish if 50.0 g of copper is formed?
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CELL EMF UNDER NONSTANDARD CONDITIONS (section 20.6) 20.61 (a) Under what circumstances is the Nernst equation applicable? (b) What is the numerical value of the reaction quotient, Q, under standard conditions? (c) What happens to the emf of a cell if the concentrations of the reactants are increased? 20.62 (a) A voltaic cell is constructed with all reactants and products in their standard states. Will this condition hold as the cell operates? Explain. (b) Can the Nernst equation be used at temperatures other than room temperature? Explain. (c) What happens to the emf of a cell if the concentrations of the products are increased?
20.67 A voltaic cell utilizes the following reaction:
20.63 What is the effect on the emf of the cell shown in Figure 20.9, which has the overall reaction Zn(s) + 2 H + (aq) ¡ Zn2 + (aq) + H2(g), for each of the following changes? (a) The pressure of the H2 gas is increased in the cathode half-cell. (b) Zinc nitrate is added to the anode half-cell. (c) Sodium hydroxide is added to the cathode half-cell, decreasing [H + ]. (d) The surface area of the anode is doubled.
(a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when [Fe 3 + ] = 3.50 M, PH2 = 0.95 atm, [Fe 2 + ] = 0.0010 M, and the pH in both half-cells is 4.00? 20.69 A voltaic cell is constructed with two Zn2 + -Zn electrodes. The two half-cells have [Zn2 + ] = 1.8 M and [Zn2 + ] = 1.00 * 10 - 2 M, respectively. (a) Which electrode is the anode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [Zn2 + ] will increase, decrease, or stay the same as the cell operates. 20.70 A voltaic cell is constructed with two silver–silver chloride electrodes, each of which is based on the following halfreaction:
20.64 A voltaic cell utilizes the following reaction: Al(s) + 3 Ag + (aq) ¡ Al3 + (aq) + 3 Ag(s) What is the effect on the cell emf of each of the following changes? (a) Water is added to the anode half-cell, diluting the solution. (b) The size of the aluminum electrode is increased. (c) A solution of AgNO3 is added to the cathode half-cell, increasing the quantity of Ag + but not changing its concentration. (d) HCl is added to the AgNO3 solution, precipitating some of the Ag + as AgCl. 20.65 A voltaic cell is constructed that uses the following reaction and operates at 298 K: Zn(s) + Ni2 + (aq) ¡ Zn2 + (aq) + Ni(s) (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when [Ni2 + ] = 3.00 M and [Zn2 + ] = 0.100 M? (c) What is the emf of the cell when [Ni2 + ] = 0.200 M and [Zn2 + ] = 0.900 M? 20.66 A voltaic cell utilizes the following reaction and operates at 298 K:
4 Fe 2 + (aq) + O2(g) + 4 H + (aq) ¡ 4 Fe 3 + (aq) + 2 H2O(l) (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when [Fe 2 + ] = 1.3 M, [Fe 3 + ] = 0.010 M, PO2 = 0.50 atm, and the pH of the solution in the cathode half-cell is 3.50? 20.68 A voltaic cell utilizes the following reaction: 2 Fe 3 + (aq) + H2(g) ¡ 2 Fe 2 + (aq) + 2 H + (aq)
AgCl(s) + e - ¡ Ag(s) + Cl - (aq) The two half-cells have [Cl - ] = 0.0150 M and [Cl - ] = 2.55 M, respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [Cl - ] will increase, decrease, or stay the same as the cell operates. 20.71 The cell in Figure 20.9 could be used to provide a measure of the pH in the cathode half-cell. Calculate the pH of the cathode half-cell solution if the cell emf at 298 K is measured to be +0.684 V when [Zn2 + ] = 0.30 M and PH2 = 0.90 atm. 20.72 A voltaic cell is constructed that is based on the following reaction:
3 Ce 4 + (aq) + Cr(s) ¡ 3 Ce 3 + (aq) + Cr 3 + (aq)
Sn2 + (aq) + Pb(s) ¡ Sn(s) + Pb 2 + (aq)
(a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when [Ce 4 + ] = 3.0 M, [Ce 3 + ] = 0.10 M, and [Cr 3 + ] = 0.010 M? (c) What is the emf of the cell when [Ce 4 + ] = 0.010 M, [Ce 3 + ] = 2.0 M, and [Cr 3 + ] = 1.5 M?
(a) If the concentration of Sn2 + in the cathode half-cell is 1.00 M and the cell generates an emf of +0.22 V, what is the concentration of Pb 2 + in the anode half-cell? (b) If the anode half-cell contains [SO42 - ] = 1.00 M in equilibrium with PbSO4(s), what is the Ksp of PbSO4?
BATTERIES AND FUEL CELLS (section 20.7) 20.73 (a) What happens to the emf of a battery as it is used? Why does this happen? (b) The AA-size and D-size alkaline batteries are both 1.5-V batteries that are based on the same electrode reactions. What is the major difference between the two batteries? What performance feature is most affected by this difference? 20.74 (a) Suggest an explanation for why liquid water is needed in an alkaline battery. (b) What is the advantage of using highly concentrated or solid reactants in a voltaic cell? 20.75 During a period of discharge of a lead-acid battery, 402 g of Pb from the anode is converted into PbSO4(s). (a) What mass of
PbO2(s) is reduced at the cathode during this same period? (b) How many coulombs of electrical charge are transferred from Pb to PbO2? 20.76 During the discharge of an alkaline battery, 4.50 g of Zn is consumed at the anode of the battery. (a) What mass of MnO2 is reduced at the cathode during this discharge? (b) How many coulombs of electrical charge are transferred from Zn to MnO2? 20.77 Heart pacemakers are often powered by lithium–silver chromate “button” batteries. The overall cell reaction is 2 Li(s) + Ag 2CrO4(s) ¡ Li2CrO4(s) + 2 Ag(s)
Exercises (a) Lithium metal is the reactant at one of the electrodes of the battery. Is it the anode or the cathode? (b) Choose the two half-reactions from Appendix E that most closely approximate the reactions that occur in the battery. What standard emf would be generated by a voltaic cell based on these half-reactions? (c) The battery generates an emf of +3.5 V. How close is this value to the one calculated in part (b)? (d) Calculate the emf that would be generated at body temperature, 37 °C. How does this compare to the emf you calculated in part (b)? 20.78 Mercuric oxide dry-cell batteries are often used where a highenergy density is required, such as in watches and cameras. The two half-cell reactions that occur in the battery are HgO(s) + H2O(l) + 2 e - ¡ Hg(l) + 2 OH - (aq) Zn(s) + 2 OH - (aq) ¡ ZnO(s) + H2O(l) + 2 e ° for the (a) Write the overall cell reaction. (b) The value of E red cathode reaction is +0.098 V. The overall cell potential is +1.35 V. Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reac-
871
tion different than would be expected if the reaction occurred in an acidic medium? 20.79 (a) Suppose that an alkaline battery was manufactured using cadmium metal rather than zinc. What effect would this have on the cell emf? (b) What environmental advantage is provided by the use of nickel-metal-hydride batteries over nickel-cadmium batteries? 20.80 (a) The nonrechargeable lithium batteries used for photography use lithium metal as the anode. What advantages might be realized by using lithium rather than zinc, cadmium, lead, or nickel? (b) The rechargeable lithium-ion battery does not use lithium metal as an electrode material. Nevertheless, it still has a substantial advantage over nickel-based batteries. Suggest an explanation. 20.81 The hydrogen-oxygen fuel cell has a standard emf of 1.23 V. What advantages and disadvantages there to using this device as a source of power compared to a 1.55-V alkaline battery? 20.82 (a) What is the difference between a battery and a fuel cell? (b) Can the “fuel” of a fuel cell be a solid? Explain.
CORROSION (section 20.8) 20.83 (a) Write the anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II). (b) Write the balanced half-reactions involved in the air oxidation of Fe 2 + (aq) to Fe 2O3 # 3 H2O.
20.86 An iron object is plated with a coating of cobalt to protect against corrosion. Does the cobalt protect iron by cathodic protection? Explain.
20.84 (a) Based on standard reduction potentials, would you expect copper metal to oxidize under standard conditions in the presence of oxygen and hydrogen ions? (b) When the Statue of Liberty was refurbished, Teflon spacers were placed between the iron skeleton and the copper metal on the surface of the statue. What role do these spacers play?
20.87 A plumber’s handbook states that you should not connect a brass pipe directly to a galvanized steel pipe because electrochemical reactions between the two metals will cause corrosion. The handbook recommends you use instead an insulating fitting to connect them. Brass is a mixture of copper and zinc. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations.
20.85 (a) Magnesium metal is used as a sacrificial anode to protect underground pipes from corrosion. Why is the magnesium referred to as a “sacrificial anode”? (b) Looking in Appendix E, suggest what metal the underground pipes could be made from in order for magnesium to be successful as a sacrificial anode.
20.88 A plumber’s handbook states that you should not connect a copper pipe directly to a steel pipe because electrochemical reactions between the two metals will cause corrosion. The handbook recommends you use instead an insulating fitting to connect them. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations.
ELECTROLYSIS (section 20.9) 20.89 (a) What is electrolysis? (b) Are electrolysis reactions thermodynamically spontaneous? Explain. (c) What process occurs at the anode in the electrolysis of molten NaCl? (d) Why is sodium metal not obtained when an aqueous solution of NaCl undergoes electrolysis? 20.90 (a) What is an electrolytic cell? (b) The negative terminal of a voltage source is connected to an electrode of an electrolytic cell. Is the electrode the anode or the cathode of the cell? Explain. (c) The electrolysis of water is often done with a small amount of sulfuric acid added to the water. What is the role of the sulfuric acid? (d) Why are active metals such as Al obtained by electrolysis using molten salts rather than aqueous solutions? 20.91 (a) A Cr 3 + (aq) solution is electrolyzed, using a current of 7.60 A. What mass of Cr(s) is plated out after 2.00 days? (b) What amperage is required to plate out 0.250 mol Cr from a Cr 3 + solution in a period of 8.00 h? 20.92 Metallic magnesium can be made by the electrolysis of molten MgCl2. (a) What mass of Mg is formed by passing a current of 4.55 A through molten MgCl2, for 4.50 days? (b) How many minutes are needed to plate out 25.00 g Mg from molten MgCl2 using 3.50 A of current?
20.93 (a) Calculate the mass of Li formed by electrolysis of molten LiCl by a current of 7.5 * 104 A flowing for a period of 24 h. Assume the electrolytic cell is 85% efficient. (b) What is the minimum voltage required to drive the reaction? 20.94 Elemental calcium is produced by the electrolysis of molten CaCl2. (a) What mass of calcium can be produced by this process if a current of 7.5 * 103 A is applied for 48 h? Assume that the electrolytic cell is 68% efficient. (b) What is the minimum voltage needed to cause the electrolysis? 20.95 Metallic gold is collected from below the anode when crude copper metal is refined by electrolysis. Explain this behavior. 20.96 The crude copper that is subjected to electrorefining contains tellurium as an impurity. The standard reduction potential between tellurium and its lowest common oxidation state, Te 4 + , is ° = 0.57 V Te 4 + (aq) + 4 e - ¡ Te(s) E red
Given this information, describe the probable fate of tellurium impurities during electrorefining. Do the impurities fall to the bottom of the refining bath, unchanged, as copper is oxidized, or do they go into solution? If they go into solution, do they plate out on the cathode?
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CHAPTER 20
Electrochemistry
ADDITIONAL EXERCISES 20.97 A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) Ni + (aq) ¡ Ni2 + (aq) + Ni(s) (acidic solution) 2(b) MnO4 (aq) ¡ MnO4 (aq) + MnO2(s) (acidic solution) (c) H2SO3(aq) ¡ S(s) + HSO4 - (aq) (acidic solution) (d) Cl 2(aq) ¡ Cl - (aq) + ClO - (aq) (basic solution) 20.98 This oxidation-reduction reaction in acidic solution is spontaneous: 5 Fe 2 + (aq) + MnO4 - (aq) + 8 H + (aq) ¡ 5 Fe 3 + (aq) + Mn2 + (aq) + 4 H2O(l) A solution containing KMnO4 and H2SO4 is poured into one beaker, and a solution of FeSO4 is poured into another. A salt bridge is used to join the beakers. A platinum foil is placed in each solution, and a wire that passes through a voltmeter connects the two solutions. (a) Sketch the cell, indicating the anode and the cathode, the direction of electron movement through the external circuit, and the direction of ion migrations through the solutions. (b) Sketch the process that occurs at the atomic level at the surface of the anode. (c) Calculate the emf of the cell under standard conditions. (d) Calculate the emf of the cell at 298 K when the concentrations are the following: pH = 0.0, [Fe 2 + ] = 0.10 M, [MnO4 - ] = 1.50 M, [Fe 3 + ] = 2.5 * 10 - 4 M, [Mn2 + ] = 0.001 M. 20.99 A common shorthand way to represent a voltaic cell is anode|anode solution||cathode solution|cathode A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by Fe|Fe 2 +||Ag +|Ag; sketch the cell. (b) Write the half-reactions and overall cell reaction represented by Zn|Zn2 +||H +|H2; sketch the cell. (c) Using the notation just described, represent a cell based on the following reaction: -
+
C lO3 (aq) + 3 Cu(s) + 6 H (aq) ¡ Cl - (aq) + 3 Cu2 + (aq) + 3 H2O(l) Pt is used as an inert electrode in contact with the ClO3 - and Cl - . Sketch the cell. 20.100 Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of Sn to Sn2 + by I2 (to form I - ), (b) reduction of Ni2 + to Ni by I (to form I2), (c) reduction of Ce 4 + to Ce 3 + by H2O2, (d) reduction of Cu2 + to Cu by Sn2 + (to form Sn4 + ). [20.101] Gold exists in two common positive oxidation states, +1 and +3. The standard reduction potentials for these oxidation states are ° = +1.69 V Au + (aq) + e - ¡ Au(s) E red 3+ ° = +1.50 V Au (aq) + 3 e ¡ Au(s) E red
(a) Can you use these data to explain why gold does not tarnish in the air? (b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction 4 Au(s) + 8 NaCN(aq) + 2 H2O(l) + O2(g) ¡ 4 Na[Au(CN)2](aq) + 4 NaOH(aq)
What is being oxidized and what is being reduced in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with Zn dust to get gold metal. Write a balanced redox reaction for this process. What is being oxidized, and what is being reduced? 20.102 A voltaic cell is constructed from an Ni2 + (aq)-Ni(s) half-cell and an Ag + (aq)-Ag(s) half-cell. The initial concentration of Ni2 + (aq) in the Ni2 + -Ni half-cell is [Ni2 + ] = 0.0100 M. The initial cell voltage is +1.12 V. (a) By using data in Table 20.1, calculate the standard emf of this voltaic cell. (b) Will the concentration of Ni2 + (aq) increase or decrease as the cell operates? (c) What is the initial concentration of Ag + (aq) in the Ag + -Ag half-cell? [20.103] A voltaic cell is constructed that uses the following half-cell reactions: Cu + (aq) + e - ¡ Cu(s) I2 (s) + 2 e - ¡ 2 I - (aq) The cell is operated at 298 K with [Cu + ] = 0.25 M and [I - ] = 3.5 M. (a) Determine E for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) If [Cu + ] was equal to 0.15 M, at what concentration of I - would the cell have zero potential? 20.104 Using data from Appendix E, calculate the equilibrium constant for the disproportionation of the copper(I) ion at room temperature: 2 Cu + (aq) ¡ Cu2 + (aq) + Cu(s). 20.105 (a) Write the reactions for the discharge and charge of a nickelcadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: Cd(OH)2(s) + 2 e - ¡ Cd(s) + 2 OH - (aq) ° = -0.76 V E red NiO(OH)(s) + H2O(l) + e - ¡ Ni(OH)2(s) + OH - (aq) ° = +0.49 V E red (c) A typical nicad voltaic cell generates an emf of +1.30 V. Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value. 20.106 The capacity of batteries such as the typical AA alkaline battery is expressed in units of milliamp-hours (mAh). An AA alkaline battery yields a nominal capacity of 2850 mAh. (a) What quantity of interest to the consumer is being expressed by the units of mAh? (b) The starting voltage of a fresh alkaline battery is 1.55 V. The voltage decreases during discharge and is 0.80 V when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge. 20.107 If you were going to apply a small potential to a steel ship resting in the water as a means of inhibiting corrosion, would you apply a negative or a positive charge? Explain. [20.108] (a) How many coulombs are required to plate a layer of chromium metal 0.25 mm thick on an auto bumper with a total area of 0.32 m2 from a solution containing CrO42 - ? The density of chromium metal is 7.20 g>cm3. (b) What current flow is required for this electroplating if the bumper is to be plated in 10.0 s? (c) If the external source has an emf of +6.0 V and the electrolytic cell is 65% efficient, how much electrical power is expended to electroplate the bumper?
Integrative Exercises 20.109 Magnesium is obtained by electrolysis of molten MgCl2. (a) Why is an aqueous solution of MgCl2 not used in the electrolysis? (b) Several cells are connected in parallel by very large copper buses that convey current to the cells. Assuming that the cells are 96% efficient in producing the desired products in electrolysis, what mass of Mg is formed by passing a current of 97,000 A for a period of 24 hr? 20.110 Calculate the number of kilowatt-hours of electricity required to produce 1.0 * 103 kg (1 metric ton) of aluminum by electrolysis of Al3+ if the applied voltage is 4.50 V and the process is 45% efficient. 20.111 Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship using a
873
surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about 7 * 108 mol of H2 to provide the buoyancy to lift the ship (J. Chem. Educ., 1973, Vol. 50, 61). (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate H2 and O2 if the pressure on the gases at the depth of the wreckage (2 mi) is 300 atm? (c) What is the minimum electrical energy required to raise the Titanic by electrolysis? (d) What is the minimum cost of the electrical energy required to generate the necessary H2 if the electricity costs 85 cents per kilowatt-hour to generate at the site?
INTEGRATIVE EXERCISES 20.112 The Haber process is the principal industrial route for converting nitrogen into ammonia: N2(g) + 3 H2(g) ¡ 2 NH3(g) (a) What is being oxidized, and what is being reduced? (b) Using the thermodynamic data in Appendix C, calculate the equilibrium constant for the process at room temperature. (c) Calculate the standard emf of the Haber process at room temperature. [20.113] In a galvanic cell the cathode is an Ag + (1.00 M)/Ag(s) halfcell. The anode is a standard hydrogen electrode immersed in a buffer solution containing 0.10 M benzoic acid (C6H5COOH) and 0.050 M sodium benzoate (C6H5COO - Na + ). The measured cell voltage is 1.030 V. What is the pKa of benzoic acid? 20.114 Consider the general oxidation of a species A in solution: A ¡ A+ + e - . The term oxidation potential is sometimes used to describe the ease with which species A is oxidized— the easier a species is to oxidize, the greater its oxidation potential. (a) What is the relationship between the standard oxidation potential of A and the standard reduction potential of A+ ? (b) Which of the metals listed in Table 4.5 has the highest standard oxidation potential? Which has the lowest? (c) For a series of substances, the trend in oxidation potential is often related to the trend in the first ionization energy. Explain why this relationship makes sense. 20.115 A voltaic cell is based on Ag + (aq)>Ag(s) and Fe 3 + (aq)> Fe 2 + (aq) half-cells. (a) What is the standard emf of the cell? (b) Which reaction occurs at the cathode and which at the anode of the cell? (c) Use S° values in Appendix C and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or decreases when the temperature is raised above 25 °C. 20.116 Hydrogen gas has the potential as a clean fuel in reaction with oxygen. The relevant reaction is 2 H2(g) + O2(g) ¡ 2 H2O(l) Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to drive a generator, much as coal is currently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at 85 °C. (a) Use data in Appendix C to calcu-
late ¢H° and ¢S° for the reaction. We will assume that these values do not change appreciably with temperature. (b) Based on the values from part (a), what trend would you expect for the magnitude of ¢G for the reaction as the temperature increases? (c) What is the significance of the change in the magnitude of ¢G with temperature with respect to the utility of hydrogen as a fuel? (d) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to generate electrical energy from hydrogen? 20.117 Cytochrome, a complicated molecule that we will represent as CyFe 2 + , reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions. (Section 19.7) At pH 7.0 the following reduction potentials pertain to this oxidation of CyFe 2 + : O2(g) + 4 H + (aq) + 4 e - ¡ 2 H2O(l) CyFe
3+
-
(aq) + e ¡ CyFe
2+
° = +0.82 V E red ° = +0.22 V E red
(aq) 2+
(a) What is ¢G for the oxidation of CyFe by air? (b) If the synthesis of 1.00 mol of ATP from adenosine diphosphate (ADP) requires a ¢G of 37.7 kJ, how many moles of ATP are synthesized per mole of O2? [20.118] The standard potential for the reduction of AgSCN(s) is +0.0895 V. AgSCN(s) + e - ¡ Ag(s) + SCN - (aq) Using this value and the electrode potential for Ag + (aq), calculate the Ksp for AgSCN. [20.119] The Ksp value for PbS(s) is 8.0 * 10 - 28. By using this value together with an electrode potential from Appendix E, determine the value of the standard reduction potential for the reaction PbS(s) + 2 e - ¡ Pb(s) + S2 - (aq) [20.120] A student designs an ammeter (a device that measures electrical current) that is based on the electrolysis of water into hydrogen and oxygen gases. When electrical current of unknown magnitude is run through the device for 2.00 min, 12.3 mL of water-saturated H2(g) is collected. The temperature of the system is 25.5 °C, and the atmospheric pressure is 768 torr. What is the magnitude of the current in amperes?
WHAT’S AHEAD 21.1 RADIOACTIVITY
21.3 NUCLEAR TRANSMUTATIONS
In this chapter we learn how to describe nuclear reactions by equations analogous to chemical equations, in which the nuclear charges and masses of reactants and products are in balance. Radioactive nuclei most commonly decay by emission of alpha, beta, or gamma radiation.
We study nuclear transmutations, which are nuclear reactions induced by bombardment of a nucleus by a neutron or an accelerated charged particle.
21.2 PATTERNS OF NUCLEAR STABILITY We recognize that nuclear stability is determined largely by the neutron-to-proton ratio. For stable nuclei, this ratio increases with increasing atomic number. All nuclei with 84 or more protons are radioactive. Heavy nuclei gain stability by a series of nuclear disintegrations leading to stable nuclei.
21.4 RATES OF RADIOACTIVE DECAY We learn that radioisotope decays are first-order kinetic processes with characteristic half-lives. Decay rates can be used to determine the age of ancient artifacts and geological formations.
21
THE REMNANT OF THE SUPERNOVA CASSIOPEIA A as viewed from the Chandra X-ray Observatory.
21.5 DETECTION OF RADIOACTIVITY We see that the radiation emitted by a radioactive substance can be detected by dosimeters, Geiger counters, and scintillation counters.
21.6 ENERGY CHANGES IN NUCLEAR REACTIONS We recognize that energy changes in nuclear reactions are related to mass changes via Einstein’s equation, E = mc 2 . The nuclear binding energy of a nucleus is the difference between the mass of the nucleus and the sum of the masses of its nucleons.
21.7 NUCLEAR POWER: FISSION We learn that in nuclear fission a heavy nucleus splits to form two or more product nuclei. This type of nuclear reaction is the energy
source for nuclear power plants, and we look at the operating principles of these plants.
21.8 NUCLEAR POWER: FUSION We learn that in nuclear fusion two light nuclei are fused together to form a more stable, heavier nucleus.
21.9 RADIATION IN THE ENVIRONMENT AND LIVING SYSTEMS We discover that naturally occurring radioisotopes bathe our planet—and us—with low levels of radiation. The radiation emitted in nuclear reactions can damage living cells but also has diagnostic and therapeutic applications.
NUCLEAR CHEMISTRY that fuels life on Earth comes from sunlight. Plants convert the radiant energy of the Sun into chemical energy through photosynthesis, producing oxygen and carbohydrates. Life on Earth could not exist without energy from the Sun, but where does the Sun get its energy? Stars, ULTIMATELY THE CHEMICAL ENERGY
including our sun, use nuclear reactions that involve changes in atomic nuclei to generate their energy. For example, the Sun produces energy by fusing hydrogen atoms to form helium, releasing vast amounts of energy in the process. The fusion of hydrogen to form helium is the dominant nuclear reaction for most of a star’s lifetime. Toward the end of its life, the hydrogen in the star’s core is exhausted and the helium atoms fuse to form progressively heavier elements. A select few stars end their lives in dramatic supernova explosions such as the one shown in the chapter opening photograph. The nuclear reactions that occur when a star goes supernova are responsible for the existence of all naturally occurring elements heavier than nickel. Nuclear chemistry is the study of nuclear reactions, with an emphasis on their uses in chemistry and their effects on biological systems. Nuclear chemistry affects our lives in many ways, particularly in energy and medical applications. In radiation therapy, for example, gamma rays from a radioactive substance such as cobalt-60 are directed to cancerous tumors to destroy them. Positron emission tomography (PET) is one example of a medical diagnostic tool that relies on decay of a radioactive element injected into the body. 875
876
CHAPTER 21
Nuclear Chemistry Oil 6%
Other 2%
Nuclear 15%
Coal 41%
Hydroelectric 16%
Gas 20%
Worldwide
Oil 1%
Other 2%
Hydroelectric Nuclear 11% 19% Coal Gas 5% 4% Hydroelectric 6%
Nuclear 77% 씰 FIGURE 21.1 Sources of electricity generation, worldwide and for select countries.
Oil 2%
Other 3%
Coal 49%
Hydroelectric 15% Nuclear 2% Gas 1% Coal 82%
Gas 21% France
United States
China
Radioactivity is also used to help determine the mechanisms of chemical reactions, to trace the movement of atoms in biological systems and the environment, and to date historical artifacts. Nuclear reactions are also used to generate electricity. Roughly 15% of the electricity generated worldwide comes from nuclear power plants, though the percentage varies from one country to the next, as 쑿 FIGURE 21.1 shows. The use of nuclear energy for power generation and the disposal of nuclear wastes from power plants are controversial social and political issues. It is imperative, therefore, that as a citizen with a stake in these matters, you have some understanding of nuclear reactions and the uses of radioactive substances.
21.1 | RADIOACTIVITY To understand nuclear reactions, we must review and develop some ideas introduced in Section 2.3. First, recall that two types of subatomic particles reside in the nucleus: protons and neutrons. We will refer to these particles as nucleons. Recall also that all atoms of a given element have the same number of protons; this number is the element’s atomic number. The atoms of a given element can have different numbers of neutrons, however, so they can have different mass numbers; the mass number is the total number of nucleons in the nucleus. Atoms with the same atomic number but different mass numbers are known as isotopes. The different isotopes of an element are distinguished by their mass numbers. For example, the three naturally occurring isotopes of uranium are uranium-234, uranium235, and uranium-238, where the numerical suffixes represent the mass numbers. These 235 238 isotopes are also written 234 92 U, 92 U, and 92 U, where the superscript is the mass number and the subscript is the atomic number. Different isotopes of an element have different natural abundances. For example, 99.3% of naturally occurring uranium is uranium-238, 0.7% is uranium-235, and only a trace is uranium-234. Different isotopes of an element also have different stabilities. Indeed, the nuclear properties of any given isotope depend on the number of protons and neutrons in its nucleus. A nuclide is a nucleus containing a specified number of protons and neutrons. Nuclides that are radioactive are called radionuclides, and atoms containing these nuclei are called radioisotopes.
SECTION 21.1
Nuclear Equations Most nuclei in nature are stable and remain intact indefinitely. Radionuclides, however, are unstable and spontaneously emit particles and electromagnetic radiation. Emission of radiation is one of the ways in which an unstable nucleus is transformed into a more stable one that has less energy. The emitted radiation is the carrier of the excess energy. Uranium-238, for example, is radioactive, undergoing a nuclear reaction emitting helium-4 nuclei. The helium-4 particles are known as alpha (a) particles, and a stream of them is called alpha radiation. When a 238 92U nucleus loses an alpha particle, the remaining fragment has an atomic number of 90 and a mass number of 234. The element with atomic number 90 is Th, thorium. Therefore, the products of uranium-238 decomposition are an alpha particle and a thorium-234 nucleus. We represent this reaction by the nuclear equation 238 92 U
¡
+ 42He
234 90 Th
[21.1]
When a nucleus spontaneously decomposes in this way, it is said either to have decayed or to have undergone radioactive decay. Because an alpha particle is involved in this reaction, scientists also describe the process as alpha decay. GIVE IT SOME THOUGHT What change in the mass number of a nucleus occurs when the nucleus emits an alpha particle?
In Equation 21.1 the sum of the mass numbers is the same on both sides of the equation (238 = 234 + 4). Likewise, the sum of the atomic numbers on both sides of the equation is equal (92 = 90 + 2). Mass numbers and atomic numbers must be balanced in all nuclear equations. The radioactive properties of the nucleus in an atom are independent of the chemical state of the atom. In writing nuclear equations, therefore, we are not concerned with the chemical form (element or compound) of the atom in which the nucleus resides. SAMPLE EXERCISE 21.1
Predicting the Product of a Nuclear Reaction
What product is formed when radium-226 undergoes alpha decay? SOLUTION Analyze We are asked to determine the nucleus that results when radium-226 loses an alpha particle. Plan We can best do this by writing a balanced nuclear reaction for the process. Solve The periodic table shows that radium has an atomic number of 88. The complete chemical symbol for radium-226 is therefore 226 88 Ra. An alpha particle is a helium-4 nucleus, and so its symbol is 42He (sometimes written as 42a ). The alpha particle is a product of the nuclear reaction, and so the equation is of the form 226 88 Ra
¡ AZX + 42He
where A is the mass number of the product nucleus and Z is its atomic number. Mass numbers and atomic numbers must balance, so 226 = A + 4 and 88 = Z + 2 Hence, A = 222
and
Z = 86
Again, from the periodic table, the element with Z = 86 is radon (Rn). The product, therefore, is 222 86 Rn , and the nuclear equation is 226 88 Ra
¡
222 86 Rn
+ 42He
PRACTICE EXERCISE Which element undergoes alpha decay to form lead-208? Answer: 212 84 Po
Radioactivity
877
878
CHAPTER 21
Nuclear Chemistry TABLE 21.1 • Properties of Alpha, Beta, and Gamma Radiation Type of Radiation Property
A
B
G
Charge Mass Relative penetrating power Nature of radiation
2+ 6.64 * 10-24 g 1 4 2He nuclei
19.11 * 10-28 g 100 Electrons
0 0 10,000 High-energy photons
Types of Radioactive Decay The three most common kinds of radiation given off when a radionuclide decays are alpha (a), beta (b), and gamma (g) radiation. • (Section 2.2) 쑿 TABLE 21.1 summarizes some of the important properties of these types of radiation. As just described, alpha radiation consists of a stream of helium-4 nuclei known as alpha particles, which we denote as 42He or 42a. Beta radiation consists of streams of beta (b) particles, which are high-speed electrons emitted by an unstable nucleus. Beta particles are represented in nuclear equations by -10e or sometimes by -10 b . The superscript 0 indicates that the mass of the electron is exceedingly small relative to the mass of a nucleon. The subscript -1 represents the negative charge of the beta particle, which is opposite that of the proton. Iodine-131 is an isotope that undergoes decay by beta emission: 131 53I
¡
131 54Xe
+
0 -1e
[21.2]
You can see from this equation that beta decay causes the atomic number of the reactant to increase from 53 to 54, which means a proton was created. Therefore, beta emission is equivalent to the conversion of a neutron (10n) to a proton (10p or 10H): 1 0n
¡ 11p +
0 -1e
[21.3]
Just because an electron is emitted from a nucleus in beta decay, we should not think that the nucleus is composed of these particles any more than we consider a match to be composed of sparks simply because it gives them off when struck. The beta-particle electron comes into being only when the nucleus undergoes a nuclear reaction. Furthermore, the speed of the beta particle is sufficiently high that it does not end up in an orbital of the decaying atom. Gamma (g) radiation (or gamma rays) consists of high-energy photons (that is, electromagnetic radiation of very short wavelength). It changes neither the atomic number nor the mass number of a nucleus and is represented as either 00g or merely g. Gamma radiation usually accompanies other radioactive emission because it represents the energy lost when the nucleons in a nuclear reaction reorganize into more stable arrangements. Generally, gamma rays are not shown when writing nuclear equations. Two other types of radioactive decay are positron emission and electron capture. A positron, 01e, is a particle that has the same mass as an electron (thus, we use the letter e and superscript 0 for the mass) but the opposite charge (represented by the +1 subscript).* The isotope carbon-11 decays by positron emission: 11 6C
¡
11 5B
+ 01e
[21.4]
Positron emission causes the atomic number of the reactant in this equation to decrease from 6 to 5. In general, positron emission has the effect of converting a proton to a neutron, thereby decreasing the atomic number of the nucleus by 1: 1 1p
¡ 10n + 01e
[21.5]
*The positron has a very short life because it is annihilated when it collides with an electron, producing gamma rays: 01e + -10 e ¡ 2 00g.
SECTION 21.1
Electron capture is the capture by the nucleus of an electron from the electron cloud surrounding the nucleus, as in this rubidium-81 decay: 81 37Rb
+
0 -1e
(orbital electron) ¡
81 36Kr
[21.6]
Because the electron is consumed rather than formed in the process, it is shown on the reactant side of the equation. Electron capture, like positron emission, has the effect of converting a proton to a neutron: +
1 1p
0 -1e
¡ 10n
[21.7]
씰 TABLE 21.2 summarizes the symbols used to represent the particles commonly encountered in nuclear reactions. The various types of radioactive decay are summarized in 쑼 TABLE 21.3.
TABLE 21.3 • Types of Radioactive Decay Type
Nuclear Equation
Alpha decay Beta emission Positron emission Electron capture*
A ZX A ZX A ZX A ZX
Change in Atomic Number
¡ AZ -- 24Y + 42He ¡ Z +A1 Y + -10e ¡ Z -A1 Y + 01e + -10 e ¡ Z -A1 Y
-2 +1 -1 -1
Change in Mass Number -4 Unchanged Unchanged Unchanged
*The electron captured comes from the electron cloud surrounding the nucleus.
GIVE IT SOME THOUGHT Which particles in Table 21.2 result in no change in nuclear charge when emitted in nuclear decay? SAMPLE EXERCISE 21.2
Writing Nuclear Equations
Write nuclear equations for (a) mercury-201 undergoing electron capture; (b) thorium-231 decaying to protactinium-231. SOLUTION Analyze We must write balanced nuclear equations in which the masses and charges of reactants and products are equal. Plan We can begin by writing the complete chemical symbols for the nuclei and decay particles that are given in the problem. Solve (a) The information given in the question can be summarized as 201 80Hg
+
0 -1e
A ZX
¡
The mass numbers must have the same sum on both sides of the equation: 201 + 0 = A Thus, the product nucleus must have a mass number of 201. Similarly, balancing the atomic numbers gives 80 - 1 = Z Thus, the atomic number of the product nucleus must be 79, which identifies it as gold (Au): 201 80Hg
+
0 -1e
¡
201 79Au
(b) In this case we must determine what type of particle is emitted in the course of the radioactive decay: 231 90 Th
¡
231 91 Pa
+ AZ X
From 231 = 231 + A and 90 = 91 + Z, we deduce A = 0 and Z = -1. According to Table 21.2, the particle with these characteristics is the beta particle (electron). We therefore write 231 90 Th
¡
231 91 Pa
+
0 -1e
Radioactivity
879
TABLE 21.2 • Particles Found in Nuclear Reactions Particle
Symbol
Neutron Proton Electron Alpha particle Beta particle Positron
1 0n 1 1 1H or 1p 0 -1e 4 4 2He or 2a 0 0 -1e or -1 b 0 1e
880
CHAPTER 21
Nuclear Chemistry
PRACTICE EXERCISE Write a balanced nuclear equation for the reaction in which oxygen-15 undergoes positron emission. Answer: 158 O ¡ 157N + 01e
21.2 | PATTERNS OF NUCLEAR STABILITY No single rule allows us to predict whether a particular nucleus is radioactive and, if it is, how it might decay. However, several empirical observations can help us predict the stability of a nucleus.
Neutron-to-Proton Ratio Because like charges repel each other, it may seem surprising that a large number of protons can reside within the small volume of the nucleus. At close distances, however, a strong force of attraction, called the nuclear force, exists between nucleons. Neutrons are intimately involved in this attractive force. All nuclei other than 11H contain neutrons. As the number of protons in a nucleus increases, there is an ever greater need for neutrons to counteract the proton–proton repulsions. Stable nuclei with atomic numbers up to about 20 have approximately equal numbers of neutrons and protons. For nuclei with atomic number above 20, the number of neutrons exceeds the number of protons. Indeed, the number of neutrons necessary to create a stable nucleus increases more rapidly than the number of protons. Thus, the neutron-toproton ratios of stable nuclei increase with increasing atomic number, as illustrated by the most common isotopes of carbon, 126C (n>p = 1), manganese, 55 25Mn (n>p = 1.20), (n>p = 1.49) Au and gold, 197 . 79 The dark blue dots in 씰 FIGURE 21.2 represent stable (nonradioactive) isotopes. The region of the graph covered by these dark blue dots is known as the belt of stability. The belt of stability ends at element 83 (bismuth), which means that all nuclei with 84 or more protons are radioactive. For example, all isotopes of uranium, Z = 92, are radioactive. The type of radioactive decay that a particular radionuclide undergoes depends largely on how its neutron-to-proton ratio compares with those of nearby nuclei that lie within the belt of stability. We can envision three general situations: 1. Nuclei above the belt of stability (high neutron-to-proton ratios). These neutron-rich nuclei can lower their ratio and thereby move toward the belt of stability by emitting a beta particle because beta emission decreases the number of neutrons and increases the number of protons (Equation 21.3). 2. Nuclei below the belt of stability (low neutron-to-proton ratios). These proton-rich nuclei can increase their ratio and so move closer to the belt of stability by either positron emission or electron capture because both decays increase the number of neutrons and decrease the number of protons (Equations 21.5 and 21.7). Positron emission is more common among lighter nuclei. Electron capture becomes increasingly common as the nuclear charge increases. 3. Nuclei with atomic numbers » 84. These heavy nuclei tend to undergo alpha emission, which decreases both the number of neutrons and the number of protons by two, moving the nucleus diagonally toward the belt of stability. SAMPLE EXERCISE 21.3
Predicting Modes of Nuclear Decay
Predict the mode of decay of (a) carbon-14, (b) xenon-118. SOLUTION Analyze We are asked to predict the modes of decay of two nuclei. Plan To do this, we must locate the respective nuclei in Figure 21.2 and determine their positions with respect to the belt of stability in order to predict the most likely mode of decay.
SECTION 21.2
Patterns of Nuclear Stability
881
Solve (a) Carbon is element 6. Thus, carbon-14 has 6 protons and 14 - 6 = 8 neutrons, giving it a neutron-to-proton ratio of 1.25. Elements with Z 6 20 normally have stable nuclei that contain approximately equal numbers of neutrons and protons (n>p = 1). Thus, carbon-14 is located above the belt of stability and we expect it to decay by emitting a beta particle to lessen the n>p ratio: 14 6C
¡
+
0 -1e
14 7N
This is indeed the mode of decay observed for carbon-14, a reaction that lowers the n>p ratio from 1.25 to 1.0. (b) Xenon is element 54. Thus, xenon-118 has 54 protons and 118 - 54 = 64 neutrons, giving it an n>p ratio of 1.18. According to Figure 21.2, stable nuclei in this region of the belt of stability have higher neutron-to-proton ratios than xenon-118. The nucleus can increase this ratio by either positron emission or electron capture: 118 54 Xe
¡ 01e +
+ -10e ¡ In this case both modes of decay are observed. 118 54 Xe
118 53 I
118 53 I
Comment Keep in mind that our guidelines do not always work. For example, thorium-233, which we might expect to undergo alpha decay, actually undergoes beta emission. Furthermore, 148 a few radioactive nuclei lie within the belt of stability. Both 146 60 Nd and 60 Nd, for example, are stable and lie in the belt of stability. 147 , however, which lies between them, is radioactive. Nd 60 PRACTICE EXERCISE Predict the mode of decay of (a) plutonium-239, (b) indium-120. Answers: (a) a decay, (b) b emission
GO FIGURE
Estimate the optimal number of neutrons for a nucleus containing 70 protons. 160 150
3
Nuclei with Z ⱖ 84, dominant decay mode ⫽ alpha emission
2
Nuclei below belt of stability, dominant decay mode ⫽ positron emission or electron capture
209 83 Bi
140
(n/p ⫽ 1.52)
130 1
120
Nuclei above belt of stability, dominant decay mode ⫽ beta emission
Number of neutrons, n
110 100 127 53 I
90
(n/p ⫽ 1.40) 80 70 60
1:1 neutronto-proton ratio
50 40 30 56 26 Fe
20
(n/p ⫽ 1.15) 16 8O
10 0
(n/p ⫽ 1.00) 0
10
20
30 40 50 60 70 Number of protons, p
80
90
100
씱 FIGURE 21.2 Stable and radioactive isotopes as a function of numbers of neutrons and protons in a nucleus. The stable nuclei (dark blue dots) define a region known as the belt of stability.
882
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Nuclear Chemistry Each blue arrow represents decay by alpha emission
Each red arrow represents decay by beta emission
238
U
236 Th
234
Pa
U
232 Th
230 228 Ra
Mass number
226 224 Rn
222 220 Po
218 216 214
Pb
Bi
Po
Pb
Bi
Po
212 210 208 206 씰 FIGURE 21.3 Nuclear disintegration series for uranium-238. The decay continues until the stable nucleus 206 82 Pb is formed.
Pb
204 81 82 83 Tl Pb Bi
84 85 86 87 88 89 90 Po At Rn Fr Ra Ac Th Atomic number
91 92 93 Pa U Np
Radioactive Series Some nuclei cannot gain stability by a single emission. Consequently, a series of successive emissions occurs as shown for uranium-238 in 쑿 FIGURE 21.3. Decay continues until a stable nucleus—lead-206 in this case—is formed. A series of nuclear reactions that begins with an unstable nucleus and terminates with a stable one is known as a radioactive series or a nuclear disintegration series. Three such series occur in nature: uranium-238 to lead-206, uranium-235 to lead-207, and thorium-232 to lead-208.
Further Observations Two further observations can help us to predict stable nuclei:
TABLE 21.4 • Number of Stable Isotopes with Even and Odd Numbers of Protons and Neutrons Number of Stable Isotopes
Proton Number
Neutron Number
157 53 50 5
Even Even Odd Odd
Even Odd Even Odd
• Nuclei with the magic numbers of 2, 8, 20, 28, 50, or 82 protons or 2, 8, 20, 28, 50, 82, or 126 neutrons are generally more stable than nuclei that do not contain these numbers of nucleons. • Nuclei with even numbers of protrons, neutrons, or both are more likely to be stable than those with odd numbers of protons and/or neutrons. Approximately 60% of stable nuclei have an even number of both protons and neutrons, whereas less than 2% have odd numbers of both (씱 TABLE 21.4). These observations can be understood in terms of the shell model of the nucleus, in which nucleons are described as residing in shells analogous to the shell structure for electrons in atoms. Just as certain numbers of electrons correspond to stable filled-shell electron configurations, so also the magic numbers of nucleons represent filled shells in nuclei. There are several examples of the stability of nuclei with magic numbers of nucleons. For example, the radioactive series in Figure 21.3 ends with the stable 206 82 Pb nucleus, which has a magic number of protons (82). Another example is the observation that tin, which has a magic number of protons (50), has ten stable isotopes, more than any other element.
SECTION 21.2
Patterns of Nuclear Stability
GO FIGURE
Among the elements shown here, how many have an even number of protons and fewer than three stable isotopes? How many have an odd number of protons and more than two stable isotopes? 1 H (2)
Number of stable isotopes 5 B (2)
6 C (2)
7 N (2)
8 O (3)
9 F (1)
2 He (2) 10 Ne (3)
13 Al (1)
14 Si (3)
15 P (1)
16 S (4)
17 Cl (2)
18 Ar (3)
30 Zn (5)
31 Ga (2)
32 Ge (4)
33 As (1)
34 Se (5)
35 Br (2)
36 Kr (6)
48 Cd (6)
49 In (1)
50 Sn (10)
51 Sb (2)
52 Te (6)
53 I (1)
54 Xe (9)
Elements with two or fewer isotopes
4 Be (1)
3 Li (2) 11 Na (1)
12 Mg (3)
19 K (2)
20 Ca (5)
21 Sc (1)
22 Ti (5)
23 V (2)
24 Cr (4)
25 Mn (1)
26 Fe (4)
27 Co (1)
28 Ni (5)
29 Cu (2)
37 Rb (1)
38 Sr (3)
39 Y (1)
40 Zr (4)
41 Nb (1)
42 Mo (6)
43 Tc (0)
44 Ru (7)
45 Rh (1)
46 Pd (6)
47 Ag (2)
Elements with three or more isotopes
쑿 FIGURE 21.4 Number of stable isotopes for elements 1–54.
Evidence also suggests that pairs of protons and pairs of neutrons have a special stability, analogous to the pairs of electrons in molecules. This evidence accounts for the second observation noted earlier, that stable nuclei with an even number of protons and/or neutrons are far more numerous than those with odd numbers. The preference for even numbers of protons is illustrated in 쑿 FIGURE 21.4, which shows the number of stable isotopes for all elements up to Xe. Notice that once we move past nitrogen, the elements with an odd number of protons invariably have fewer stable isotopes than their neighbors with an even number of protons. GIVE IT SOME THOUGHT What can you say about the number of neutrons in the stable isotopes of fluorine, sodium, aluminum, and phosphorus?
SAMPLE EXERCISE 21.4
Predicting Nuclear Stability
98 Predict which of these nuclei are especially stable: 42He, 40 20Ca, 43Tc.
SOLUTION Analyze We are asked to identify especially stable nuclei, given their mass numbers and atomic numbers. Plan We look to see whether the numbers of protons and neutrons correspond to magic numbers. Solve The 42He nucleus (the alpha particle) has a magic number of both protons (2) and neutrons (2) and is very stable. The 40 20Ca nucleus also has a magic number of both protons (20) and neutrons (20) and is especially stable. The 98 43Tc nucleus does not have a magic number of either protons or neutrons. In fact, it has an odd number of both protons (43) and neutrons (55). There are very few stable nuclei with odd numbers of both protons and neutrons. Indeed, technetium-98 is radioactive. PRACTICE EXERCISE Which of the following nuclei would you expect to exhibit a special stability: 208 82 Pb? 208 Answer: 118 50 Sn, 82 Pb
118 210 50 Sn, 85 At,
883
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Nuclear Chemistry
21.3 | NUCLEAR TRANSMUTATIONS Thus far we have examined nuclear reactions in which a nucleus decays spontaneously. A nucleus can also change identity if it is struck by a neutron or by another nucleus. Nuclear reactions induced in this way are known as nuclear transmutations. In 1919, Ernest Rutherford performed the first conversion of one nucleus into another, using alpha particles emitted by radium to convert nitrogen-14 into oxygen-17: 14 7N
+ 42He ¡
17 8O
+ 11H
[21.8]
Such reactions have allowed scientists to synthesize hundreds of radioisotopes in the laboratory. Nuclear transmutations are sometimes represented by listing, in order, the target nucleus, the bombarding particle, the ejected particle, and the product nucleus. Using this condensed notation, Equation 21.8 becomes Target nucleus
Product nucleus 14 7
N (␣, p)178 O
Bombarding particle
SAMPLE EXERCISE 21.5
Ejected particle
Writing a Balanced Nuclear Equation
24 Write the balanced nuclear equation for the process summarized as 27 13 Al(n, a)11 Na.
SOLUTION Analyze We must go from the condensed descriptive form of the reaction to the balanced nuclear equation. Plan We arrive at the balanced equation by writing n and α, each with its associated subscripts and superscripts. Solve The n is the abbreviation for a neutron (10 n) and α represents an alpha particle (42He). The neutron is the bombarding particle, and the alpha particle is a product. Therefore, the nuclear equation is 27 13Al
+ 10n ¡
24 11Na
+ 42He
PRACTICE EXERCISE Write the condensed version of the nuclear reaction Answer: 168 O(p,
a)137 N
16 8O
+ 11H ¡
13 7N
+ 42He
Accelerating Charged Particles Alpha particles and other positively charged particles must move very fast to overcome the electrostatic repulsion between them and the target nucleus. The higher the nuclear charge on either the bombarding particle or the target nucleus, the faster the bombarding particle must move to bring about a nuclear reaction. Many methods have been devised to accelerate charged particles, using strong magnetic and electrostatic fields. These particle accelerators, popularly called “atom smashers,” bear such names as cyclotron and synchrotron. A common theme in all particle accelerators is the need to create charged particles so that they can be manipulated by electrical and magnetic fields. The tubes through which the particles move must be kept at high vacuum so that the particles do not inadvertently collide with any gas-phase molecules. 씰 FIGURE 21.5 shows the Relativistic Heavy Ion Collider (RHIC) located at Brookhaven National Laboratory. This facility and the Large Hadron Collider (LHC) at CERN (Conseil Européen pour la Recherche Nucléaire) near Geneva are two of the largest particle accelerators in the world. Both LHC and RHIC are capable of accelerating protons, as well as heavy ions such as gold and lead, to speeds approaching the speed
SECTION 21.3
4
3
Nuclear Transmutations
Finally the ions are transferred to RHIC, which has a circumference of 3.8 km. Ions moving in opposite directions can collide at one of six points on the ring, marked with white rectangles
The booster synchrotron and alternate gradient synchrotron (AGS) further accelerate the ions to 99.7% of the speed of light
1
2
If needed, beams of H + ions can be generated in the Linac
쑿 FIGURE 21.5 The Relativistic Heavy Ion Collider. This particle accelerator is located at Brookhaven National Laboratory on Long Island, New York.
of light. Scientists study the outcomes of collisions involving these ultra-high-energy particles. These experiments are used to investigate the fundamental structure of matter and ultimately answer questions about the beginning of the universe.
Reactions Involving Neutrons Most synthetic isotopes used in medicine and scientific research are made using neutrons as the bombarding particles. Because neutrons are neutral, they are not repelled by the nucleus. Consequently, they do not need to be accelerated to cause nuclear reactions. The neutrons are produced in nuclear reactors. For example, cobalt-60, which is used in cancer radiation therapy, is produced by neutron capture. Iron-58 is placed in a nuclear reactor and bombarded by neutrons to trigger the reactions sequence 58 26Fe
+ 10n ¡ 59 26Fe
59 27Co
+
1 0n
59 26Fe
¡
59 27Co
¡
60 27Co
[21.9] +
0 -1e
[21.10] [21.11]
GIVE IT SOME THOUGHT Can an electrostatic or magnetic field be used to accelerate neutrons in a particle accelerator? Why or why not?
Transuranium Elements Nuclear transmutations have been used to produce the elements with atomic number above 92, collectively known as the transuranium elements because they follow
Gold atoms are ionized, creating ions that are accelerated in Tandem van de Graaff accelerator
885
886
CHAPTER 21
Nuclear Chemistry
uranium in the periodic table. Elements 93 (neptunium, Np) and 94 (plutonium, Pu) were produced in 1940 by bombarding uranium-238 with neutrons: 238 92 U
+ 10n ¡ 239 93 Np
239 92 U
¡
¡
239 94 Pu
239 93 Np
+
+
0 -1e
0 -1e
[21.12] [21.13]
Elements with still larger atomic numbers are normally formed in small quantities in particle accelerators. Curium-242, for example, is formed when a plutonium-239 target is bombarded with accelerated alpha particles: 239 94 Pu
+ 42He ¡
242 96 Cm
+ 10n
[21.14]
In 1996 a team of European scientists based in Germany synthesized element 112, copernicium, Cn, by bombarding a lead target continuously for three weeks with a beam of zinc atoms: 208 82 Pb
+
70 30Zn
¡
277 112Cn
+ 10n
[21.15]
Amazingly, their discovery was based on the detection of only one atom of the new element, which decays after roughly 100 μs by alpha decay to form darmstadtium-273 (element 110). Within one minute, another five alpha decays take place producing fermium-253 (element 100). The finding has been verified in both Japan and Russia. More recently, scientists have reported the synthesis of elements 113 through 118. These results have yet to be confirmed by the International Union for Pure and Applied Chemistry, although the results look promising. Names and symbols have not yet been chosen for these new elements.
21.4 | RATES OF RADIOACTIVE DECAY
Mass of 90 Sr (grams)
Some radioisotopes, such as uranium-238, are found in nature even though they are not stable. Other radioisotopes do not exist in nature but can be synthesized in nuclear reactions. To understand this distinction, we must realize that different nuclei undergo radioactive decay at different rates. Many radioisotopes decay essenGO FIGURE If we start with a 50.0-g sample, how much of tially completely in a matter of seconds, so we do not find them in nature. Uranium-238, on the other hand, decays very slowly. Therefore, despite its it remains after three half-lives have passed? instability, we can still observe what remains from its formation in the early history of the universe. Radioactive decay is a first-order kinetic process. Recall that a first-order process has a characteristic half-life, which is the time required for half of any given quantity of a substance to react. • (Section 14.4) Nuclear decay rates are commonly expressed in terms of half-lives. Each isotope has its own char10.0 acteristic half-life. For example, the half-life of strontium-90 is 28.8 yr (씱 FIGURE 21.6). If we start with 10.0 g of strontium-90, only 5.0 g of that 8.0 isotope remains after 28.8 yr, 2.5 g remains after another 28.8 yr, and so on. Strontium-90 decays to yttrium-90: 6.0
90 38Sr
4.0 2.0
0
20
40 60 80 Time (years)
100
120
쑿 FIGURE 21.6 Decay of a 10.0-g sample of strontium-90 (t1/2 ⴝ 28.8 yr). The 10 * 10 grids show how much of the radioactive isotope remains after various amounts of time.
¡
90 39Y
+
0 -1e
[21.16]
Half-lives as short as millionths of a second and as long as billions of years are known. The half-lives of some radioisotopes are listed in 씰 TABLE 21.5. One important feature of half-lives for nuclear decay is that they are unaffected by external conditions such as temperature, pressure, or state of chemical combination. Unlike toxic chemicals, therefore, radioactive atoms cannot be rendered harmless by chemical reaction or by any other practical treatment. At this point, we can do nothing but allow these nuclei to lose radioactivity at their characteristic rates. In the meantime, we must take precautions to prevent radioisotopes, such as those produced in nuclear power plants (Section 21.7), from entering the environment because of the damage radiation can cause.
SECTION 21.4
TABLE 21.5 • The Half-Lives and Type of Decay for Several Radioisotopes Isotope
Half-Life (yr)
Type of Decay
Natural radioisotopes
238 92 U 235 92 U 232 90 Th 40 19K 14 6C
4.5 * 7.0 * 1.4 * 1.3 * 5715
Alpha Alpha Alpha Beta Beta
Synthetic radioisotopes
239 94 Pu 137 55 Cs 90 38Sr 131 53 I
24,000 30 28.8 0.022
SAMPLE EXERCISE 21.6
109 108 1010 109
Alpha Beta Beta Beta
Calculation Involving Half-Lives
The half-life of cobalt-60 is 5.3 yr. How much of a 1.000-mg sample of cobalt-60 is left after 15.9 yr? SOLUTION Analyze We are given the half-life for cobalt-60 and asked to calculate the amount of cobalt-60 remaining from an initial 1.000-mg sample after 15.9 yr. Plan We will use the fact that the amount of a radioactive substance decreases by 50% for every half-life that passes. Solve Because 5.3 * 3 = 15.9, 15.9 yr is three half-lives for cobalt-60. At the end of one half-life, 0.500 mg of cobalt-60 remains, 0.250 mg at the end of two half-lives, and 0.125 mg at the end of three half-lives. PRACTICE EXERCISE Carbon-11, used in medical imaging, has a half-life of 20.4 min. The carbon-11 nuclides are formed, and the carbon atoms are then incorporated into an appropriate compound. The resulting sample is injected into a patient, and the medical image is obtained. If the entire process takes five half-lives, what percentage of the original carbon-11 remains at this time? Answer: 3.12%
Radiometric Dating Because the half-life of any particular nuclide is constant, the half-life can serve as a nuclear clock to determine the age of objects. The method of dating objects based on their isotopes and isotope abundances is called radiometric dating. When carbon-14 is used in radiometric dating, the technique is known as radiocarbon dating. The procedure is based on the formation of carbon-14 as neutrons created by cosmic rays in the upper atmosphere convert nitrogen-14 into carbon-14 (씰 FIGURE 21.7). The 14C reacts with oxygen to form 14CO2 in the atmosphere, and this “labeled” CO2 is taken up by plants and introduced into the food chain through photosynthesis. This process provides a small but reasonably constant source of carbon-14, which is radioactive and undergoes beta decay with a half-life of 5715 yr: 14 6C
¡
14 7N
+
0 -1e
[21.17]
Because a living plant or animal has a constant intake of carbon compounds, it is able to maintain a ratio of carbon-14 to carbon-12 that is nearly identical with that of the atmosphere. Once the organism dies, however, it no longer ingests carbon compounds to replenish the carbon-14 lost through radioactive decay. The ratio of carbon-14 to carbon-12 therefore decreases. By measuring this ratio and comparing it with that of the atmosphere, we can estimate the age of an object. For example, if the ratio diminishes to half that of the atmosphere, we can conclude that the object is one half-life, or 5715 yr old. This method cannot be used to date objects older than about 50,000 yr because after this length of time the radioactivity is too low to be measured accurately.
Rates of Radioactive Decay
887
888
CHAPTER 21
Nuclear Chemistry
1
Cosmic rays (largely protons) enter the atmosphere and collide with atoms, creating neutrons
2
Nitrogen atoms capture a neutron and emit a proton, forming 14C
3
14C atoms are incorporated in CO2, which is taken up by plants and made into more complex molecules through photosynthesis
4
Animals and people take in 14C by eating plants
5
Once an organism dies, intake of 14C ceases and its concentration decreases through beta emission to form 14N
1 0n
14 7N
14
⫹ 10 n
C ⫹ O2
14 6C
14
⫹ 11 p
CO2
씰 FIGURE 21.7 Creation and distribution of carbon-14. The ratio of carbon-14 to carbon-12 in a dead animal or plant is related to the time since death occurred.
14 6C
14 7N
⫹ ⫺10e
In radiocarbon dating, a reasonable assumption is that the ratio of carbon-14 to carbon-12 in the atmosphere has been relatively constant for the past 50,000 yr. However, because variations in solar activity control the amount of carbon-14 produced in the atmosphere, that ratio can fluctuate. We can correct for this effect by using other kinds of data. Recently scientists have compared carbon-14 data with data from tree rings, corals, lake sediments, ice cores, and other natural sources to correct variations in the carbon-14 “clock” back to 26,000 yr. Other isotopes can be similarly used to date other types of objects. For example, it takes 4.5 * 109 yr for half of a sample of uranium-238 to decay to lead-206. The age of rocks containing uranium can therefore be determined by measuring the ratio of lead206 to uranium-238. If the lead-206 had somehow become incorporated into the rock by normal chemical processes instead of by radioactive decay, the rock would also contain large amounts of the more abundant isotope lead-208. In the absence of large amounts of this “geonormal” isotope of lead, it is assumed that all of the lead-206 was at one time uranium-238. The oldest rocks found on Earth are approximately 3 * 109 yr old. This age indicates that Earth’s crust has been solid for at least this length of time. Scientists estimate that it required 1 * 109 to 1.5 * 109 yr for Earth to cool and its surface to become solid, making the age of Earth 4.0 to 4.5 * 109 yr.
Calculations Based on Half-Life So far, our discussion has been mainly qualitative. We now consider the topic of halflives from a more quantitative point of view. This approach enables us to determine the half-life of a radioisotope or the age of an object. As noted earlier, radioactive decay is a first-order kinetic process. Its rate, therefore, is proportional to the number of radioactive nuclei N in a sample: Rate = kN
[21.18]
The first-order rate constant, k, is called the decay constant. The rate at which a sample decays is called its activity, and it is often expressed as number of disintegrations per unit time. The becquerel (Bq) is the SI unit for expressing activity. A becquerel is defined as one nuclear disintegration per second. An older,
SECTION 21.4
but still widely used, unit of activity is the curie (Ci), defined as 3.7 * 1010 disintegrations per second, which is the rate of decay of 1 g of radium. Thus, a 4.0-mCi sample of cobalt-60 undergoes 3.7 * 1010 disintegrations/s = 1.5 * 108 disintegrations/s 1 Ci and so has an activity of 1.5 * 108 Bq. As a radioactive sample decays, the amount of radiation emanating from the sample decays as well. For example, the half-life of cobalt-60 is 5.26 yr. The 4.0-mCi sample of cobalt-60 would, after 5.26 yr, have a radiation activity of 2.0 mCi, or 7.5 * 107 Bq. 4.0 * 10-3 Ci *
GIVE IT SOME THOUGHT Why can’t spontaneous radioactive decay be a zero-order or second-order kinetic process?
As we saw in Section 14.4, a first-order rate law can be transformed into the equation ln
Nt = -kt N0
[21.19]
In this equation t is the time interval of decay, k is the decay constant, N0 is the initial number of nuclei (at time zero), and Nt is the number remaining after the time interval. Both the mass of a particular radioisotope and its activity are proportional to the number of radioactive nuclei. Thus, either the ratio of the mass at any time t to the mass at time t = 0 or the ratio of the activities at time t and t = 0 can be substituted for Nt>N0 in Equation 21.19. From Equation 21.19 we can obtain the relationship between the decay constant, k, and half-life, t1/2: • (Section 14.4) k =
0.693 t1/2
[21.20]
where we have used the value ln (Nt>N0) = ln (0.5) = -0.693 for one half-life. Thus, if we know the value of either the decay constant or the half-life, we can calculate the value of the other. GIVE IT SOME THOUGHT a. Would doubling the mass of a radioactive sample change the activity of the sample? b. Would doubling the mass change the half-life for the radioactive decay? SAMPLE EXERCISE 21.7
Calculating the Age of a Mineral
A rock contains 0.257 mg of lead-206 for every milligram of uranium-238. The half-life for the decay of uranium-238 to lead-206 is 4.5 * 109 yr. How old is the rock? SOLUTION Analyze We are told that a rock sample has a certain amount of lead-206 for every unit mass of uranium-238 and asked to estimate the age of the rock. Plan Lead-206 is the product of the radioactive decay of uranium-238. We will assume that the only source of lead-206 in the rock is from the decay of uranium-238, with a known halflife. To apply first-order kinetics expressions (Equations 21.19 and 21.20) to calculate the time elapsed since the rock was formed, we first need to calculate how much initial uranium-238 there was for every 1 mg that remains today. Solve Let’s assume that the rock currently contains 1.000 mg of uranium-238 and therefore 0.257 mg of lead-206. The amount of uranium-238 in the rock when it was first formed therefore equals 1.000 mg plus the quantity that has decayed to lead-206. Because the mass of lead atoms is not the same as the mass of uranium atoms, we cannot just add 1.000 mg and 0.257 mg. We have to multiply the present mass of lead-206 (0.257 mg) by the ratio of the mass number of uranium to that of lead, into which it has decayed. Therefore, the original mass of 238 92 U was Original 238 92 U = 1.000 mg + = 1.297 mg
238 (0.257 mg) 206
Rates of Radioactive Decay
889
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CHAPTER 21
Nuclear Chemistry
Using Equation 21.20, we can calculate the decay constant for the process from its half-life: k =
0.693 = 1.5 * 10-10 yr -1 4.5 * 109 yr
Rearranging Equation 21.19 to solve for time, t, and substituting known quantities gives t = -
Nt 1 1 1.000 = ln ln = 1.7 * 109 yr k N0 1.5 * 10-10 yr -1 1.297
Comment To check this result, you could use the fact that the decay of uranium-237 to lead-207 has a half-life of 7 * 108 yr and measure the relative amounts of uranium-237 and lead-207 in the rock. PRACTICE EXERCISE A wooden object from an archeological site is subjected to radiocarbon dating. The activity due to 14C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second. The half-life of 14C is 5715 yr. What is the age of the archeological sample? Answer: 2230 yr
We can also use measurements of radioactive activity to determine the half-life of a radioactive isotope, as the next sample exercise shows. SAMPLE EXERCISE 21.8
Calculations Involving Radioactive Decay
If we start with 1.000 g of strontium-90, 0.953 g will remain after 2.00 yr. (a) What is the half-life of strontium-90? (b) How much strontium-90 will remain after 5.00 yr? (c) What is the initial activity of the sample in becquerels and curies? SOLUTION (a) Analyze We are asked to calculate a half-life, t1/2, based on data that tell us how much of a radioactive nucleus has decayed in a time interval t = 2.00 yr and the information N0 = 1.000 g, Nt = 0.953 g. Plan We first calculate the rate constant for the decay, k, and then use that to compute t1>2. Solve Equation 21.19 is solved for the decay constant, k, and then Equation 21.20 is used to calculate half-life, t1>2: k = = t1>2 =
0.953 g Nt 1 1 ln ln = t N0 2.00 yr 1.000 g 1 (-0.0481) = 0.0241 yr -1 2.00 yr
0.693 0.693 = = 28.8 yr k 0.0241 yr -1
(b) Analyze We are asked to calculate the amount of a radionuclide remaining after a given period of time. Plan We need to calculate Nt, the amount of strontium present at time t, using the initial quantity, N0, and the rate constant for decay, k, calculated in part (a). Solve Again using Equation 21.19, with k = 0.0241 yr -1, we have ln
Nt = -kt = -(0.0241 yr -1)(5.00 yr) = -0.120 N0
Nt>N0 is calculated from ln(Nt>N0) = -0.120 using the ex or INV LN function of a calculator: Nt = e -0.120 = 0.887 N0 Because N0 = 1.000 g, we have Nt = (0.887)N0 = (0.887)(1.000 g) = 0.887 g (c) Analyze We are asked to calculate the activity of the sample in becquerels and curies. Plan We must calculate the number of disintegrations per atom per second and then multiply by the number of atoms in the sample.
SECTION 21.5
Detection of Radioactivity
891
Solve The number of disintegrations per atom per second is given by the rate constant, k: k = a
1 yr 1 day 1h 0.0241 ba ba ba b = 7.64 * 10-10 s -1 yr 365 days 24 h 3600 s
To obtain the total number of disintegrations per second, we calculate the number of atoms in the sample. We multiply this quantity by k, where we express k as the number of disintegrations per atom per second, to obtain the number of disintegrations per second: (1.000 g 90Sr)a
1 mol 90Sr 6.022 * 1023 atoms Sr ba b = 6.7 * 1021 atoms 90Sr 90 g 90Sr 1 mol 90Sr
Total disintegrations/s = a
7.64 * 10-10 disintegrations b (6.7 * 1021 atoms) atom # s
= 5.1 * 1012 disintegrations>s Because 1 Bq is one disintegration per second, the activity is 5.1 * 1012 Bq. The activity in curies is given by (5.1 * 1012 disintegrations/s)a
1 Ci b = 1.4 * 102 Ci 3.7 * 1010 disintegrations>s
We have used only two significant figures in products of these calculations because we do not know the atomic weight of 90Sr to more than two significant figures without looking it up in a special source. PRACTICE EXERCISE A sample to be used for medical imaging is labeled with 18F, which has a half-life of 110 min. What percentage of the original activity in the sample remains after 300 min? Answer: 15.1%
21.5 | DETECTION OF RADIOACTIVITY A variety of methods have been devised to detect emissions from radioactive substances. Henri Becquerel discovered radioactivity because radiation caused fogging of photographic plates, and since that time photographic plates and film have been used to detect radioactivity. The radiation affects photographic film in much the same way as X-rays do. The greater the extent of exposure to radiation, the darker the area of the developed negative. People who work with radioactive substances carry film badges to record the extent of their exposure to radiation (쑼 FIGURE 21.8). Radioactivity can also be detected and measured by a Geiger counter. The operation of this device is based on the fact that radiation is able to ionize matter. The ions and electrons produced by the ionizing radiation permit conduction of an electrical current. The basic design of a Geiger counter is shown in 씰 FIGURE 21.9. A current pulse between the anode and the metal cylinder occurs whenever entering radiation produces ions. Each pulse is counted in order to estimate the amount of radiation.
The film strip is white before exposure to radiation
The film strip is darkened on exposure to radiation
씱 FIGURE 21.8 Badge dosimeters monitor the extent to which the individual has been exposed to high-energy radiation. The radiation dose is determined from the extent of darkening of the film in the dosimeter.
892
CHAPTER 21
Nuclear Chemistry
4
Current is amplified and measured as series of pulses, with each pulse signaling detection of a radioactive particle or ray.
Metal cylinder acting as cathode (⫺)
1
Anode (⫹)
Radiation (α-, β-, or γ- rays) penetrates thin window
Argon gas e⫺ 3 9 8 0 0 0
Ar⫹
Amplifier and counter
␥-ray High voltage
씰 FIGURE 21.9 Schematic drawing of a Geiger counter.
3
Charged particles moving between anode and cathode create electric current.
Thin window penetrated by radiation 2
Radiation ionizes gaseous atoms (usually Ar or He), creating positively charged ions and electrons
GIVE IT SOME THOUGHT Will alpha, beta, and gamma rays pass through the window of a Geiger counter with equal efficiency?
Substances that are electronically excited by radiation can also be used to detect and measure radiation. For example, some substances excited by radiation give off light as electrons return to their lower-energy states. These substances are called phosphors. Different substances respond to different particles. Zinc sulfide, for example, responds to alpha particles. An instrument called a scintillation counter is used to detect and measure radiation, based on the tiny flashes of light produced when radiation strikes a suitable phosphor. The flashes of light are magnified electronically and counted to measure the amount of radiation.
Radiotracers Because radioisotopes can be detected readily, they can be used to follow an element through its chemical reactions. The incorporation of carbon atoms from CO2 into glucose during photosynthesis, for example, has been studied using CO2 enriched in carbon-14: Sunlight
6 14CO2 + 6 H2O ——— ¡ Chlorophyll
C6H12O6 + 6 O2
14
[21.21]
The use of the carbon-14 label provides direct experimental evidence that carbon dioxide in the environment is chemically converted to glucose in plants. Analogous labeling experiments using oxygen-18 show that the O2 produced during photosynthesis comes from water, not carbon dioxide. When it is possible to isolate and purify intermediates and products from reactions, detection devices such as scintillation counters can be used to “follow” the radioisotope as it moves from starting material through intermediates to final product. These types of experiments are useful for identifying elementary steps in a reaction mechanism. • (Section 14.6) The use of radioisotopes is possible because all isotopes of an element have essentially identical chemical properties. When a small quantity of a radioisotope is mixed with the naturally occurring stable isotopes of the same element, all the isotopes go through the same reactions together. The element’s path is revealed by the radioactivity of the radioisotope. Because the radioisotope can be used to trace the path of the element, it is called a radiotracer.
SECTION 21.5
Detection of Radioactivity
893
CHEMISTRY AND LIFE MEDICAL APPLICATIONS OF RADIOTRACERS Radiotracers have found wide use as diagnostic tools in medicine. 쑼 TABLE 21.6 lists some radiotracers and their uses. These radioisotopes are incorporated into a compound that is administered to the patient, usually intravenously. The diagnostic use of these isotopes TABLE 21.6 • Some Radionuclides Used as Radiotracers Nuclide
Half-Life
Area of the Body Studied
Iodine-131 Iron-59 Phosphorus-32 Technetium-99 Thallium-201 Sodium-24
8.04 days 44.5 days 14.3 days 6.0 hours 73 hours 14.8 hours
Thyroid Red blood cells Eyes, liver, tumors Heart, bones, liver, and lungs Heart, arteries Circulatory system
Scintillation counters detect gamma rays
Gamma rays moving in opposite directions are created when a positron and an electron collide, annihilating each other 쑿 FIGURE 21.10 Schematic representation of a positron emission tomography (PET) scanner.
is based on the ability of the radioactive compound to localize and concentrate in the organ or tissue under investigation. Iodine-131, for example, has been used to test the activity of the thyroid gland. This gland is the only important user of iodine in the body. The patient drinks a solution of NaI containing iodine-131. Only a very small amount is used so that the patient does not receive a harmful dose of radioactivity. A Geiger counter placed close to the thyroid, in the neck region, determines the ability of the thyroid to take up the iodine. A normal thyroid will absorb about 12% of the iodine within a few hours. The medical applications of radiotracers are further illustrated by positron emission tomography (PET). PET is used for clinical diagnosis of many diseases. In this method, compounds containing radionuclides that decay by positron emission are injected into a patient. These compounds are chosen to enable researchers to monitor blood flow, oxygen and glucose metabolic rates, and other biological functions. Some of the most interesting work involves the study of the brain, which depends on glucose for most of its energy. Changes in how this sugar is metabolized or used by the brain may signal a disease such as cancer, epilepsy, Parkinson’s disease, or schizophrenia. The compound to be detected in the patient must be labeled with a radionuclide that is a positron emitter. The most widely used nuclides are carbon-11 (t1>2 = 20.4 min), fluorine-18 (t1>2 = 110 min), oxygen-15 (t1>2 = 2 min), and nitrogen13 (t1>2 = 10 min). Glucose, for example, can be labeled with carbon-11. Because the half-lives of positron emitters are so short, the chemist must quickly incorporate the radionuclide into the sugar (or other appropriate) molecule and inject the compound immediately. The patient is placed in an instrument that measures the positron emission and constructs a computer-based image of the organ in which the emitting compound is localized. When the element decays, the emitted positron quickly collides with an electron. The positron and electron are annihilated in the collision, producing two gamma rays that move in opposite directions. The gamma rays are detected by an encircling ring of scintillation counters Radioactive isotope (씱 FIGURE 21.10). Because the rays decays emitting move in opposite directions but were crea positron ated in the same place at the same time, it is possible to accurately locate the point in the body where the radioactive isotope decayed. The nature of this image provides clues to the presence of disease or other abnormality and helps medical researchers understand how a particular disease affects the functioning of the brain. For example, the images shown in 씱 FIGURE 21.11 reveal that levels of activity in brains of patients with Alzheimer’s disease are different from the levels in those without the disease. RELATED EXERCISES: 21.53, 21.54
씱 FIGURE 21.11 Positron emission tomography (PET) scans showing glucose metabolism levels in the brain. Red and yellow colors show higher levels of glucose metabolism.
894
CHAPTER 21
Nuclear Chemistry
|
21.6 ENERGY CHANGES IN NUCLEAR REACTIONS The energies associated with nuclear reactions can be considered with the aid of Einstein’s celebrated equation relating mass and energy: E = mc 2
[21.22]
In this equation E stands for energy, m for mass, and c for the speed of light, 2.9979 * 108 m/s. This equation states that the mass and energy of an object are proportional. If a system loses mass, it loses energy (exothermic); if it gains mass, it gains energy (endothermic). Because the proportionality constant in the equation, c 2, is such a large number, even small changes in mass are accompanied by large changes in energy. The mass changes in chemical reactions are too small to detect. For example, the mass change associated with the combustion of 1 mol of CH4 (an exothermic process) is -9.9 * 10-9 g. Because the mass change is so small, it is possible to treat chemical reactions as though mass is conserved. The mass changes and the associated energy changes in nuclear reactions are much greater than those in chemical reactions. The mass change accompanying the radioactive decay of 1 mol of uranium-238, for example, is 50,000 times greater than that for the combustion of 1 mol of CH4. Let’s examine the energy change for the nuclear reaction 238 92 U
¡
234 90 Th
+ 42He
234 4 The masses of the nuclei are 238 92 U, 238.0003 amu; 90 Th, 233.9942 amu; and 2 He, 4.0015 amu. The mass change, ¢m, is the total mass of the products minus the total mass of the reactants. The mass change for the decay of 1 mol of uranium-238 can then be expressed in grams:
233.9942 g + 4.0015 g - 238.0003 g = -0.0046 g The fact that the system has lost mass indicates that the process is exothermic. All spontaneous nuclear reactions are exothermic. The energy change per mole associated with this reaction is ¢E = ¢(mc 2) = c 2 ¢m = (2.9979 * 108 m/s)2(-0.0046 g)a = -4.1 * 1011
kg-m2 s2
1 kg b 1000 g
= -4.1 * 1011 J
Notice that ¢m must be converted to kilograms, the SI unit of mass, to obtain ¢E in joules, the SI unit of energy. The negative sign for the energy change indicates that energy is released in the reaction—in this case, over 400 billion joules per mole of uranium!
SAMPLE EXERCISE 21.9
Calculating Mass Change in a Nuclear Reaction
How much energy is lost or gained when 1 mol of cobalt-60 undergoes beta decay, 60 0 60 60 60 27Co ¡ -1e + 28Ni? The mass of a 27Co atom is 59.933819 amu, and that of a 28Ni atom is 59.930788 amu. SOLUTION Analyze We are asked to calculate the energy change in a nuclear reaction. Plan We must first calculate the mass change in the process. We are given atomic masses, but we need the masses of the nuclei in the reaction. We calculate these by taking account of the masses of the electrons that contribute to the atomic masses. -4 Solve A 60 amu. (See the 27Co atom has 27 electrons. The mass of an electron is 5.4858 * 10 list of fundamental constants in the back inside cover.) We subtract the mass of the 27 elec60 trons from the mass of the 60 27Co atom to find the mass of the 27Co nucleus:
59.933819 amu - (27)(5.4858 * 10-4 amu) = 59.919007 amu (or 59.919007 g/mol)
SECTION 21.6
Energy Changes in Nuclear Reactions
Likewise, for 60 28Ni, the mass of the nucleus is 59.930788 amu - (28)(5.4858 * 10-4 amu) = 59.915428 amu (or 59.915428 g/mol) The mass change in the nuclear reaction is the total mass of the products minus the mass of the reactant: 60 ¢m = mass of electron + mass 60 28Ni nucleus - mass of 27Co nucleus
= 0.00054858 amu + 59.915428 amu - 59.919007 amu = -0.003030 amu Thus, when a mole of cobalt-60 decays, ¢m = - 0.003030 g Because the mass decreases (¢m 6 0), energy is released (¢E 6 0). The quantity of energy released per mole of cobalt-60 is calculated using Equation 21.22: ¢E = c 2 ¢m = (2.9979 * 108 m/s)2(-0.003030 g)a = -2.723 * 1011
kg-m2 s2
1 kg b 1000 g
= -2.723 * 1011 J
PRACTICE EXERCISE 11 Positron emission from 11C, 116C ¡ 115B + 01e, occurs with release of 2.87 * 10 J per mole 11 11 of C. What is the mass change per mole of C in this nuclear reaction? The masses of 11B and 11C are 11.009305 and 11.011434 amu, respectively. Answer: -3.19 * 10-3 g
Nuclear Binding Energies Scientists discovered in the 1930s that the masses of nuclei are always less than the masses of the individual nucleons of which they are composed. For example, the helium-4 nucleus has a mass of 4.00150 amu. The mass of a proton is 1.00728 amu and that of a neutron is 1.00866 amu. Consequently, two protons and two neutrons have a total mass of 4.03188 amu: Mass of two protons = 2(1.00728 amu) = 2.01456 amu Mass of two neutrons = 2(1.00866 amu) = 2.01732 amu Total mass = 4.03188 amu The mass of the individual nucleons is 0.03038 amu greater than that of the helium-4 nucleus: Mass of two protons and two neutrons = 4.03188 amu Mass of 42He nucleus = 4.00150 amu Mass difference ¢m = 0.03038 amu The mass difference between a nucleus and its constituent nucleons is called the mass defect. The origin of the mass defect is readily understood if we consider that energy must be added to a nucleus to break it into separated protons and neutrons: Energy + 42He ¡ 2 11p + 2 10n
[21.23]
The addition of energy to a system must be accompanied by a proportional increase in mass. The mass change we just calculated for the conversion of helium-4 into separated nucleons is ¢m = 0.03038 amu. Therefore, the energy required for this process is ¢E = c 2 ¢m = (2.9979 * 108 m/s)2(0.03038 amu) a
1g 6.022 * 10 amu 23
ba
1 kg b 1000 g
= 4.534 * 10-12 J The energy required to separate a nucleus into its individual nucleons is called the nuclear binding energy. The mass defect and nuclear binding energy for three elements are compared in 씰 TABLE 21.7.
895
896
CHAPTER 21
Nuclear Chemistry TABLE 21.7 • Mass Defects and Binding Energies for Three Nuclei Nucleus
Mass of Nucleus (amu)
Mass of Individual Nucleons (amu)
Mass Defect (amu)
Binding Energy (J)
Binding Energy per Nucleon (J)
4 2 He 56 26 Fe 238 92 U
4.00150 55.92068 238.00031
4.03188 56.44914 239.93451
0.03038 0.52846 1.93420
4.53 * 10-12 7.90 * 10-11 2.89 * 10-10
1.13 * 10-12 1.41 * 10-12 1.21 * 10-12
GIVE IT SOME THOUGHT The atomic mass of iron-56 is 55.93494 amu. Why is this number different from the mass of the nucleus given in Table 21.7?
Values of binding energies per nucleon can be used to compare the stabilities of different combinations of nucleons (such 4 235 He as two protons and two neutrons arranged either as 42He or as U 2 21H). 씱 FIGURE 21.12 shows average binding energy per nuSplitting nuclei cleon plotted against mass number. Binding energy per nucleon 1 (fission) at first increases in magnitude as mass number increases, reachreleases energy ing about 1.4 * 10-12 J for nuclei whose mass numbers are in the vicinity of iron-56. It then decreases slowly to about 1.2 * 10-12 J for very heavy nuclei. This trend indicates that nuCombining nuclei clei of intermediate mass numbers are more tightly bound (and (fusion) therefore more stable) than those with either smaller or larger releases energy mass numbers. 0 This trend has two significant consequences: First, heavy nuclei gain stability and therefore give off energy if they are fragmented into two midsized nuclei. This process, known as 0 50 100 150 200 250 fission, is used to generate energy in nuclear power plants. SecMass number ond, even greater amounts of energy are released if very light nuclei are combined, or fused together, to give more massive nuclei. This fusion process 쑿 FIGURE 21.12 Nuclear binding energies. The average binding energy per is the essential energy-producing process in the Sun. 56
nucleon increases initially as the mass number increases and then decreases slowly. Because of these trends, fusion of light nuclei and fission of heavy nuclei are exothermic processes.
Increased stability
Binding energy per nucleon (10⫺12 J)
Fe
GIVE IT SOME THOUGHT Could fusing two stable nuclei that have mass numbers in the vicinity of 100 be an energy-releasing process?
21.7 | NUCLEAR POWER: FISSION Commercial nuclear power plants and most forms of nuclear weaponry depend on nuclear fission for their operation. The first nuclear fission reaction to be discovered was that of uranium-235. This nucleus, as well as those of uranium-233 and plutonium-239, undergoes fission when struck by a slow-moving neutron (쑼 FIGURE 21.13).* 91 36Kr 1 0n
씰 FIGURE 21.13 Uranium-235 fission. This is just one of many fission patterns. In this reaction, 3.5 * 10-11 J of energy is produced per 235U nucleus.
235 92U
142 56Ba
1 0n
*Other heavy nuclei can be induced to undergo fission. However, these three are the only ones of practical importance.
SECTION 21.7
Nuclear Power: Fission
897
A heavy nucleus can split in many ways. Two ways that the uranium-235 nucleus splits, for instance, are 1 0n
⫹
235 92U
137 52Te
⫹
97 40Zr
⫹ 2 10 n
[21.24]
142 56Ba
⫹
91 36Kr
⫹ 3 10 n
[21.25]
More than 200 isotopes of 35 elements have been found among the fission products of uranium-235. Most of them are radioactive. Slow-moving neutrons are required in fission because the process involves initial absorption of the neutron by the nucleus. The resulting more massive nucleus is often unstable and spontaneously undergoes fission. Fast neutrons tend to bounce off the nucleus, and little fission occurs. Note that the coefficients of the product neutrons in Equations 21.24 and 21.25 are 2 and 3. On average, 2.4 neutrons are produced by every fission of a uranium-235 nucleus. If one fission produces two neutrons, the two neutrons can cause two additional fissions, each producing two neutrons. The four neutrons thereby released can produce four fissions, and so forth, as shown in 씰 FIGURE 21.14. The number of fissions and the energy released quickly escalate, and if the process is unchecked, the result is a violent explosion. Reactions that multiply in this fashion are called chain reactions. For a fission chain reaction to occur, the sample of fissionable material must have a certain minimum mass. Otherwise, neutrons escape from the sample before they have the opportunity to strike other nuclei and cause additional fission. The amount of fissionable material large enough to maintain a chain reaction with a constant rate of fission is called the critical mass. When a critical mass of material is present, one neutron on average from each fission is subsequently effective in producing another fission and the fission continues at a constant, controllable rate. The critical mass of uranium-235 is about 50 kg for a bare sphere of the metal.* If more than a critical mass of fissionable material is present, very few neutrons escape. The chain reaction thus multiplies the number of fissions, which can lead to a nuclear explosion. A mass in excess of a critical mass is referred to as a supercritical mass. The effect of mass on a fission reaction is illustrated in 쑼 FIGURE 21.15. 씰 FIGURE 21.16 shows a schematic diagram of the first atomic bomb used in warfare, the bomb dropped on Hiroshima, Japan, on August 6, 1945. To trigger a fission
Neutron in Nucleus
2 neutrons from fission
쑿 FIGURE 21.14 Fission chain reaction.
Subcritical uranium-235 target
Subcritical uranium-235 wedge Subcritical mass Rate of neutron loss ⬎ rate of neutron creation by fission
Critical mass Rate of neutron loss ⫽ rate of neutron creation by fission
Supercritical mass Rate of neutron loss ⬍ rate of neutron creation by fission
쑿 FIGURE 21.15 Subcritical, critical, and supercritical fission. *The exact value of the critical mass depends on the shape of the radioactive substance. The critical mass can be reduced if the radioisotope is surrounded by a material that reflects some neutrons.
Chemical explosive 쑿 FIGURE 21.16 An atomic bomb design. A conventional explosive is used to bring two subcritical masses together to form a supercritical mass.
898
CHAPTER 21
Nuclear Chemistry
A CLOSER LOOK THE DAWNING OF THE NUCLEAR AGE Uranium-235 fission was first achieved during the late 1930s by Enrico Fermi and coworkers in Rome and shortly thereafter by Otto Hahn and coworkers in Berlin. Both groups were trying to produce transuranium elements. In 1938, Hahn identified barium among his reaction products. He was puzzled by this observation and questioned the identification because the presence of barium was so unexpected. He sent a letter describing his experiments to Lise Meitner, a former coworker who had been forced to leave Germany because of the anti-Semitism of the Third Reich and had settled in Sweden. She surmised that Hahn’s experiment indicated a nuclear process was occurring in which the uranium-235 split. She called this process nuclear fission. Meitner passed word of this discovery to her nephew, Otto Frisch, a physicist working at Niels Bohr’s institute in Copenhagen. Frisch repeated the experiment, verifying Hahn’s observations, and found that tremendous energies were involved. In January 1939, Meitner and Frisch published a short article describing the reaction. In March 1939, Leo Szilard and Walter Zinn at Columbia University discovered that more neutrons are produced than are used in each fission. As we have seen, this result allows a chain reaction to occur. News of these discoveries and an awareness of their potential use in explosive devices spread rapidly within the scientific community. Several scientists finally persuaded Albert Einstein, the most famous physicist of the time, to write a letter to President Franklin D. Roosevelt explaining the implications of these discoveries. Einstein’s letter, written in August 1939, outlined the possible military applica-
Control-rod drive
tions of nuclear fission and emphasized the danger that weapons based on fission would pose if they were developed by the Nazis. Roosevelt judged it imperative that the United States investigate the possibility of such weapons. Late in 1941, the decision was made to build a bomb based on the fission reaction. An enormous research project, known as the Manhattan Project, began. On December 2, 1942, the first artificial self-sustaining nuclear fission chain reaction was achieved in an abandoned squash court at the University of Chicago. This accomplishment led to the development of the first atomic bomb, at Los Alamos National Laboratory in New Mexico in July 1945 (쑼 FIGURE 21.17). In August 1945 the United States dropped atomic bombs on two Japanese cities, Hiroshima and Nagasaki. The nuclear age had arrived.
쑿 FIGURE 21.17 The Trinity test for the atom bomb developed during World War II. The first human-made nuclear explosion took place on July 16, 1945, on the Alamogordo test range in New Mexico.
reaction, two subcritical masses of uranium-235 are slammed together using chemical explosives. The combined masses of the uranium form a supercritical mass, which leads to a rapid, uncontrolled chain reaction and, ultimately, a nuclear explosion. The energy released by the bomb dropped on Hiroshima was equivalent to that of 20,000 tons of TNT (it therefore is called a 20-kiloton bomb). Unfortunately, the basic design of a fission-based atomic bomb is quite simple, and the fissionable materials are potentially available to any nation with a nuclear reactor. The combination of design simplicity and materials availability has resulted in the proliferation of atomic weapons.
Nuclear Reactors Control rods
Fuel elements
Water acts as both moderator and coolant 쑿 FIGURE 21.18 Diagram of a pressurized water reactor core.
Nuclear power plants use nuclear fission to generate energy. The core of a typical nuclear reactor consists of four principal components: fuel elements, control rods, a moderator, and a primary coolant (씱 FIGURE 21.18). The fuel is a fissionable substance, such as uranium-235. The natural isotopic abundance of uranium-235 is only 0.7%, too low to sustain a chain reaction in most reactors. Therefore, the 235U content of the fuel must be enriched to 3–5% for use in a reactor. The fuel elements contain enriched uranium in the form of UO2 pellets encased in zirconium or stainless steel tubes. The control rods are composed of materials that absorb neutrons, such as cadmium or boron. These rods regulate the flux of neutrons to keep the reaction chain selfsustaining and also prevent the reactor core from overheating.* The probability that a neutron will trigger fission of a 235U nucleus depends on the speed of the neutron. The neutrons produced by fission have high speeds (typically *The reactor core cannot reach supercritical levels and explode with the violence of an atomic bomb because the concentration of uranium-235 is too low. However, if the core overheats, sufficient damage can lead to release of radioactive materials into the environment.
SECTION 21.7
Nuclear Power: Fission
GO FIGURE
Why are nuclear power plants usually located near a large body of water? 2
Heat is transferred to the secondary coolant in the heat exchanger, generating steam 3
Containment shell 1
Turbine
Pressurized water is heated in the reactor core
Steam
Condensor
The steam drives an electric generator, creating electricity
Electric generator 4
Reactor core Heat exchanger Primary coolant (H2O)
Pump
Heat is transferred to an external source of water, condensing the secondary coolant, which is pumped back to the heat exchanger
Pump Pump
27 °C
River
38 °C
Secondary coolant (H2O) 쑿 FIGURE 21.19 Basic design of a pressurized water reactor nuclear power plant.
in excess of 10,000 km>s). The function of the moderator is to slow down the neutrons (to speeds of a few kilometers per second) so that they can be captured more readily by the fissionable nuclei. The moderator is typically either water or graphite. The primary coolant is a substance that transports the heat generated by the nuclear chain reaction away from the reactor core. In a pressurized water reactor, which is the most common commercial reactor design, water acts as both the moderator and the primary coolant. The design of a nuclear power plant is basically the same as that of a power plant that burns fossil fuel (except that the burner is replaced by a reactor core). The nuclear power plant design shown in 쑿 FIGURE 21.19, a pressurized water reactor, is currently the most popular. The primary coolant passes through the core in a closed system, which lessens the chance that radioactive products could escape the core. As an added safety precaution, the reactor is surrounded by a reinforced concrete containment shell to shield personnel and nearby residents from radiation and to protect the reactor from external forces. After passing through the reactor core, the very hot primary coolant passes through a heat exchanger where much of its heat is transferred to a secondary coolant, converting the latter to high-pressure steam that is used to drive a turbine. The secondary coolant is then condensed by transferring heat to an external source of water, such as a river or lake. Approximately two-thirds of all commercial reactors are pressurized water reactors, but there are several variations on this basic design, each with advantages and disadvantages. A boiling water reactor generates steam by boiling the primary coolant; thus, no secondary coolant is needed. Pressurized water reactors and boiling water reactors are collectively referred to as light water reactors because they use H2O as moderator and primary coolant. A heavy water reactor uses D2O (D = deuterium,2 H) as moderator and primary coolant, and a gas-cooled reactor uses a gas, typically CO2, as primary coolant and graphite as the moderator. Use of either D2O or graphite as the moderator has the advantage that both substances absorb fewer neutrons than H2O. Consequently, the uranium fuel does not need to be enriched (though the reactor can also be run with enriched fuel).
899
900
CHAPTER 21
Nuclear Chemistry Graphite
Silicon carbide
Graphite shell
Porous carbon buffer
Embedded fuel particles
6 cm
UO2 fuel kernel
씰 FIGURE 21.20 Fuel spheres used in a high-temperature pebble-bed reactor. The image on the right is an optical microscope image of a fuel particle.
~1 mm Cross-sectional view
Coated fuel particle
In a high-temperature pebble-bed reactor, the fuel elements are spheres (“pebbles”) roughly the size of an orange (쑿 FIGURE 21.20). The spheres are made of graphite, which acts as the moderator, and thousands of tiny fuel particles are embedded in the interior of each sphere. Each fuel particle is a kernel of fissionable material, typically 235U in the form of UO2, surrounded by carbon and a coating of a ceramic material, such as SiC. Hundreds of thousands of these spheres are loosely packed in the reactor core, and helium gas, which acts as the primary coolant, flows up through the packed spheres. The reactor core operates at temperatures considerably higher than those in a light water reactor, approaching 950 °C. A pebble-bed reactor is not subject to steam explosions and does not need to be shut down to refuel. Engineers can remove spent spheres from the bottom of the reactor core and add fresh ones to the top. This design is relatively new and is not yet in commercial use.
Nuclear Waste The fission products that accumulate as a reactor operates decrease the efficiency of the reactor by capturing neutrons. For this reason, commercial reactors must be stopped periodically to either replace or reprocess the nuclear fuel. When the fuel elements are removed from the reactor, they are initially very radioactive. It was originally intended that they be stored for several months in pools at the reactor site to allow decay of short-lived radioactive nuclei. They were then to be transported in shielded containers to reprocessing plants where the fuel would be separated from the fission products. Reprocessing plants have been plagued with operational difficulties, however, and there is intense opposition in the United States to the transport of nuclear wastes on the nation’s roads. Even if the transportation difficulties could be overcome, the high level of radioactivity of the spent fuel makes reprocessing a hazardous operation. At present in the United States spent fuel elements are kept in storage at reactor sites. Spent fuel is reprocessed, however, in France, Russia, the United Kingdom, India, and Japan. Storage of spent nuclear fuel poses a major problem because the fission products are extremely radioactive. It is estimated that 20 half-lives are required for their radioactivity to reach levels acceptable for biological exposure. Based on the 28.8-yr half-life of strontium90, one of the longer-lived and most dangerous of the products, the wastes must be stored for 600 years. Plutonium-239 is one of the by-products present in spent fuel elements. It is formed by absorption of a neutron by uranium-238, followed by two successive beta emissions. (Remember that most of the uranium in the fuel elements is uranium-238.) If the elements are reprocessed, the plutonium-239 is largely recovered because it can be used as a nuclear fuel. However, if the plutonium is not removed, spent elements must be stored for a very long time because plutonium-239 has a half-life of 24,000 yr. One approach to getting more power out of existing uranium sources and potentially reducing radioactive waste is a fast breeder reactor. This type of reactor is so named because it creates (“breeds”) more fissionable material than it consumes. The reactor operates without a moderator, which means the neutrons used are not slowed down. In order to capture the fast neutrons, the fuel must be highly enriched with both uranium-235 and plutonium-239. Water cannot be used as a primary coolant because it
Nuclear Power: Fission
SECTION 21.7
would moderate the neutrons, and so a liquid metal, usually sodium, is used. The core is surrounded by a blanket of uranium-238 that captures neutrons that escape the core, producing plutonium-239 in the process. The plutonium can later be separated by reprocessing and used as fuel in a future cycle. Because fast neutrons are more effective at decaying many radioactive nuclides, the material separated from the uranium and plutonium during reprocessing is less radioactive than waste from other reactors. However, generation of relatively high levels of plutonium coupled with the need for reprocessing is problematic in terms of nuclear nonproliferation. Thus, political factors coupled with increased safety concerns and higher operational costs make fast breeder reactors quite rare. A considerable amount of research is being devoted to disposal of radioactive wastes. At present, the most attractive possibilities appear to be formation of glass, ceramic, or synthetic rock from the wastes, as a means of immobilizing them. These solid materials would then be placed in containers of high corrosion resistance and durability and buried deep underground. The U.S. Department of Energy (DOE) had designated Yucca Mountain in Nevada as a disposal site, and extensive construction has been done there. However, as of the writing of this book, the DOE has publicly stated that the Yucca Mountain site will not be used for storage, although some members of Congress are fighting this decision. The long-term solution to nuclear waste storage in the United States remains unclear. Whatever the solution finally decided on, there must be assurances that the solids and their containers will not crack from the heat generated by nuclear decay, allowing radioactivity to find its way into underground water supplies. In spite of all these difficulties, nuclear power is making a modest comeback as an energy source. The threat of global warming has moved some organizations to propose nuclear power as a major energy source in the future. Increasing demand for power in developing Asian countries has sparked a rise in construction of new nuclear power plants in that part of the world (쑼 FIGURE 21.21).
GO FIGURE
Which country has the most reactors in operation? Which country has the most reactors under construction? Which country generates the highest percentage of its electricity from nuclear power? Sweden United Kingdom Canada 18
19
2 15%
United States 104 1 20%
10 0
Ukraine
0
15 18% 35% Germany G France
0
Russia
49%
32
20
6
9
35%
18%
Japan 58
17
1
South Korea 54
0
75%
India
26% 19
4
2%
China 11
Percentage of country’s electricity generated by nuclear power 104 Number of reactors in operation 23 Number of reactors under construction 쑿 FIGURE 21.21 Number of reactors in operation and under construction for the countries with the largest nuclear power generation capabilities.
23
2%
2 29%
901
902
CHAPTER 21
Nuclear Chemistry
21.8 | NUCLEAR POWER: FUSION Energy is produced when light nuclei fuse into heavier ones. Reactions of this type are responsible for the energy produced by the Sun. Spectroscopic studies indicate that the mass composition of the Sun is 73% H, 26% He, and only 1% all other elements. The following reactions are among the numerous fusion processes believed to occur in the Sun: 1 1H 1 1H
+ 11H ¡ 21H + 01e
[21.26]
+
[21.27]
2 1H
3 3 2He + 2He 1 3 2He + 1H
¡ ¡ ¡
3 2He 4 2He 4 2He
+ +
2 11H 0 1e
[21.28] [21.29]
Fusion is appealing as an energy source because of the availability of light isotopes on Earth and because fusion products are generally not radioactive. Despite this fact, fusion is not presently used to generate energy. The problem is that, in order for two nuclei to fuse, high temperatures and pressures are needed to overcome the electrostatic repulsion between them. Fusion reactions are therefore also known as thermonuclear reactions. The lowest temperature required for any fusion is about 40,000,000 K, the temperature needed to fuse deuterium and tritium: 2 1H
+ 31H ¡ 42He + 10n
[21.30]
Such high temperatures have been achieved by using an atomic bomb to initiate fusion. This is the operating principle behind a thermonuclear, or hydrogen, bomb. This approach is obviously unacceptable, however, for a power generation plant.* Numerous problems must be overcome before fusion becomes a practical energy source. In addition to the high temperatures necessary to initiate the reaction, there is the problem of confining the reaction. No known structural material is able to withstand the enormous temperatures necessary for fusion. Research has centered on the use of an apparatus called a tokamak, which uses strong magnetic fields to contain and to heat the reaction. Temperatures of over 100,000,000 K have been achieved in a tokamak. Unfortunately, scientists have not yet been able to generate more power than is consumed over a sustained period of time.
|
21.9 RADIATION IN THE ENVIRONMENT AND LIVING SYSTEMS We are continuously bombarded by radiation from both natural and artificial sources. We are exposed to infrared, ultraviolet, and visible radiation from the Sun; radio waves from radio and television stations; microwaves from microwave ovens; X-rays from medical procedures; and radioactivity from natural materials (씰 TABLE 21.8). Understanding the different energies of these various kinds of radiation is necessary in order to understand their different effects on matter. When matter absorbs radiation, the radiation energy can cause atoms in the matter to be either excited or ionized. In general, radiation that causes ionization, called ionizing radiation, is far more harmful to biological systems than radiation that does not cause ionization. The latter, called nonionizing radiation, is generally of lower energy, such as radiofrequency electromagnetic radiation • (Section 6.7) or slow-moving neutrons. Most living tissue contains at least 70% water by mass. When living tissue is irradiated, water molecules absorb most of the energy of the radiation. Thus, it is common to define ionizing radiation as radiation that can ionize water, a process requiring a minimum energy of 1216 kJ>mol. Alpha, beta, and gamma rays (as well as X-rays and higher-energy ultraviolet radiation) possess energies in excess of this quantity and are therefore forms of ionizing radiation. *Historically a nuclear weapon that relies solely on a fission process to release energy is called an atomic bomb, whereas one that also releases energy via a fusion reaction is called a hydrogen bomb.
SECTION 21.9
Radiation in the Environment and Living Systems
903
A CLOSER LOOK NUCLEAR SYNTHESIS OF THE ELEMENTS The lightest elements––hydrogen and helium along with very small amounts of lithium and beryllium––were formed as the universe expanded in the moments following the Big Bang. All the heavier elements owe their existence to nuclear reactions that occur in stars. These heavier elements are not all created equally, however. Carbon and oxygen are a million times more abundant than lithium and boron, for instance, and over 100 million times more abundant that beryllium (쑼 FIGURE 21.22)! In fact, of the elements heavier than helium, carbon and oxygen are the most abundant. This is more than an academic curiosity given the fact that these elements, together with hydrogen, are the most important elements for life on Earth. Let’s look at the factors responsible for the relatively high abundance of carbon and oxygen in the universe. 1 ⫻ 100
H
He
Abundance in solar system
⫺2
1 ⫻ 10
O
C 1 ⫻ 10⫺4
A star is born from a cloud of gas and dust called a nebula. When conditions are right, gravitational forces collapse the cloud, and its core density and temperature rise until nuclear fusion commences. Hydrogen nuclei fuse to form deuterium, 2H, and eventually 4 He through the reactions shown in Equations 21.26 through 21.29. Because 4He has a larger binding energy than any of its immediate neighbors (Figure 21.12), these reactions release an enormous amount of energy. This process, called hydrogen burning, is the dominant process for most of a star’s lifetime. Once a star’s supply of hydrogen is nearly exhausted, several important changes occur as the star enters the red giant phase of its life. The decrease in nuclear fusion causes the core to contract, triggering an increase in core temperature and pressure. At the same time, the outer regions expand and cool enough to make the star emit red light (thus, the name red giant). The star now must use 42 He nuclei as its fuel. The simplest reaction that can occur in the He-rich core, fusion of two alpha particles to form a 84Be nucleus, does occur. However, this nucleus is highly unstable (half-life of 7 * 10-17 s) and so falls apart almost immediately. In a tiny fraction of cases, however, a third 4 8 2He collides with a 4Be nucleus before it decays, forming carbon-12 through the triple-alpha process:
Ne N
Some of the oxygen-16:
1 ⫻ 10⫺6
12 6C
4 2He
+ 42He ¡ 84Be
8 4Be
+ 42He ¡
nuclei go on to react with alpha particles to form 12 6C
1 ⫻ 10⫺8
12 6C
+ 42He ¡
16 8O
This stage of nuclear fusion is called helium burning. Notice that carbon, element 6, is formed without prior formation of elements 3, 4, B Li and 5, explaining in part their unusually low abundance. Nitrogen is 1 ⫻ 10⫺10 relatively abundant because it can be produced from carbon through Be a series of reactions involving proton capture and positron emission. Most stars gradually cool and dim as the helium is converted to 1 ⫻ 10⫺12 0 2 4 6 8 10 carbon and oxygen, ending their lives as white dwarfs. In stars that are Atomic number 10 or more times more massive than our Sun, however, a more dramatic fate awaits. The extreme mass of these stars leads to much higher 쑿 FIGURE 21.22 Abundances of elements 1–10 in the solar system. Note the logarithmic scale used for the y-axis. temperatures and pressures at the core, where a variety of fusion processes lead to synthesis of the elements from neon to sulfur. These fusion reactions are collectively called advanced burning. Eventually progressively heavier elements form at the core until it becomes predominantly 56Fe as shown in 씱 FIGURE Hydrogen burning (H He) 21.23. Because this is such a stable nucleus, further fusion to heavier nuclei consumes energy rather than releasing it. When this happens, the fusion reactions that power the star diminish, Helium burning and immense gravitational forces lead to a dramatic collapse (He C, O) called a supernova explosion. Neutron capture coupled with subsequent radioactive decays in the dying moments of such a star Advanced burning (Ne through S) are responsible for the presence of all elements heavier than iron and nickel. Without these dramatic supernova events, heavier ele56 26Fe core ments that are so familiar to us, such as silver, gold, iodine, lead, and uranium, would not exist. F
RELATED EXERCISES: 21.70, 21.72 씱 FIGURE 21.23 Fusion processes going on in a red giant just prior to a supernova explosion.
904
CHAPTER 21
Nuclear Chemistry
TABLE 21.8 • Average Abundances and Activities of Natural Radionuclides* Potassium-40
Rubidium-87
Thorium-232
Uranium-238 2.8
Land elemental abundance (ppm)
28,000
112
10.7
Land activity (Bq>kg)
870
102
43
Ocean elemental concentration (mg>L)
339
0.12
35
1 * 10
-7 -7
0.0032
Ocean activity (Bq>L)
12
0.11
4 * 10
Ocean sediments elemental abundance (ppm)
17,000
—
5.0
1.0
Ocean sediments activity (Bq>kg)
500
—
20
12
Human body activity (Bq)
4000
600
0.08
0.4**
0.040
*Data from “Ionizing Radiation Exposure of the Population of the United States,” Report 93, 1987, National Council on Radiation Protection. **Includes lead-210 and polonium-210, daughter nuclei of uranium-238.
When ionizing radiation passes through living tissue, electrons are removed from water molecules, forming highly reactive H2O + ions. An H2O + ion can react with another water molecule to form an H3O + ion and a neutral OH molecule: H2O + + H2O ¡ H3O + + OH
GO FIGURE
Why are alpha rays much more dangerous when the source of radiation is located inside the body? Skin Tissue
a b g
Bone
Organs
쑿 FIGURE 21.24 Relative penetrating abilities of alpha, beta, and gamma radiation.
[21.31]
The unstable and highly reactive OH molecule is a free ..radical, a substance with one or more unpaired electrons, as seen in the Lewis structure . O .. ¬ H . The OH molecule is also called the hydroxyl radical, and the presence of the unpaired electron is often emphasized by writing the species with a single dot, . OH . In cells and tissues, hydroxyl radicals can attack biomolecules to produce new free radicals, which in turn attack yet other biomolecules. Thus, the formation of a single hydroxyl radical via Equation 21.31 can initiate a large number of chemical reactions that are ultimately able to disrupt the normal operations of cells. The damage produced by radiation depends on the activity and energy of the radiation, the length of exposure, and whether the source is inside or outside the body. Gamma rays are particularly harmful outside the body because they penetrate human tissue very effectively, just as X-rays do. Consequently, their damage is not limited to the skin. In contrast, most alpha rays are stopped by skin, and beta rays are able to penetrate only about 1 cm beyond the skin surface (씱 FIGURE 21.24). Neither alpha rays nor beta rays are as dangerous as gamma rays, therefore, unless the radiation source somehow enters the body. Within the body, alpha rays are particularly dangerous because they transfer their energy efficiently to the surrounding tissue, causing considerable damage. In general, the tissues damaged most by radiation are those that reproduce rapidly, such as bone marrow, blood-forming tissues, and lymph nodes. The principal effect of extended exposure to low doses of radiation is to cause cancer. Cancer is caused by damage to the growth-regulation mechanism of cells, inducing the cells to reproduce uncontrollably. Leukemia, which is characterized by excessive growth of white blood cells, is probably the major type of radiation-caused cancer. In light of the biological effects of radiation, it is important to determine whether any levels of exposure are safe. Unfortunately, we are hampered in our attempts to set realistic standards because we do not fully understand the effects of long-term exposure. Scientists concerned with setting health standards have used the hypothesis that the effects of radiation are proportional to exposure. Any amount of radiation is assumed to cause some finite risk of injury, and the effects of high dosage rates are extrapolated to those of lower ones. Other scientists believe, however, that there is a threshold below which there are no radiation risks. Until scientific evidence enables us to settle the matter with some confidence, it is safer to assume that even low levels of radiation present some danger.
Radiation Doses Two units are commonly used to measure exposure to radiation. The gray (Gy), the SI unit of absorbed dose, corresponds to the absorption of 1 J of energy per kilogram of tissue. The rad (radiation absorbed dose) corresponds to the absorption of 1 * 10-2 J
SECTION 21.9
Radiation in the Environment and Living Systems
905
of energy per kilogram of tissue. Thus, 1 Gy = 100 rad. The rad is the unit most often used in medicine. Not all forms of radiation harm biological materials with the same efficiency. For example, 1 rad of alpha radiation can produce more damage than 1 rad of beta radiation. To correct for these differences, the radiation dose is multiplied by a factor that measures the relative damage caused by the radiation. This multiplication factor is known as the relative biological effectiveness, RBE. The RBE is approximately 1 for gamma and beta radiation, and 10 for alpha radiation. The exact value of the RBE varies with dose rate, total dose, and type of tissue affected. The product of the radiation dose in rads and the RBE of the radiation give the effective dosage in rem (roentgen equivalent for man): Number of rem = (number of rad) (RBE)
[21.32]
The SI unit for effective dose is the sievert (Sv), obtained by multiplying the RBE times the SI unit for radiation dose, the gray; because a gray is 100 times larger than a rad, 1 Sv = 100 rem. The rem is the unit of radiation damage usually used in medicine. GIVE IT SOME THOUGHT If a 50-kg person is uniformly irradiated by 0.10-J alpha radiation, what is the absorbed dosage in rad and the effective dosage in rem?
The effects of short-term exposure to radiation appear in 쑼 TABLE 21.9. An exposure of 600 rem is fatal to most humans. To put this number in perspective, a typical dental X-ray entails an exposure of about 0.5 mrem. The average exposure for a person in 1 year due to all natural sources of ionizing radiation (called background radiation) is about 360 mrem (쑼 FIGURE 21.25).
TABLE 21.9 • Effects of Short-Term Exposures to Radiation Dose (rem)
Effect
0–25 25–50 100–200 500
No detectable clinical effects Slight, temporary decrease in white blood cell counts Nausea; marked decrease in white blood cell counts Death of half the exposed population within 30 days
Average annual exposure (mrem) 0
50 Radon (200 mrem)
100
150
200
250
Rocks and soil (28 mrem) Natural sources (82%)
Cosmic rays (27 mrem) Radioisotopes in the body (40 mrem) Medical X-rays (39 mrem) Nuclear medicine (14 mrem) Consumer products (11 mrem)
Human-made sources (18%)
씱 FIGURE 21.25 Sources of U.S. average annual exposure to high-energy radiation. The total average annual exposure is 360 mrem. Data from “Ionizing Radiation Exposure of the Population of the United States,” Report 93, 1987, National Council on Radiation Protection.
906
CHAPTER 21
Nuclear Chemistry
Radon Radon-222 is a product of the nuclear disintegration series of uranium-238 (Figure 21.3) and is continuously generated as uranium in rocks and soil decays. As Figure 21.25 indicates, radon exposure is estimated to account for more than half the 360-mrem average annual exposure to ionizing radiation. The interplay between the chemical and nuclear properties of radon makes it a health hazard. Because radon is a noble gas, it is extremely unreactive and is therefore free to escape from the ground without chemically reacting along the way. It is readily inhaled and exhaled with no direct chemical effects. Its half-life, however, is only 3.82 days. It decays, by losing an alpha particle, into a radioisotope of polonium: 222 86 Rn
¡
218 84 Po
+ 42He
[21.33]
Because radon has such a short half-life and because alpha particles have a high RBE, inhaled radon is considered a probable cause of lung cancer. Even worse than the radon, however, is the decay product because polonium-218 is an alpha-emitting chemically active element that has an even shorter half-life (3.11 min) than radon-222: 218 84 Po
¡
214 82 Pb
+ 42He
[21.34]
When a person inhales radon, therefore, atoms of polonium-218 can become trapped in the lungs, where they bathe the delicate tissue with harmful alpha radiation. The resulting damage is estimated to contribute to 10% of all lung cancer deaths in the United States. The U.S. Environmental Protection Agency (EPA) has recommended that radon-222 levels not exceed 4 pCi per liter of air in homes. Homes located in areas where the natural uranium content of the soil is high often have levels much greater than that (쑼 FIGURE 21.26). Because of public awareness, radon-testing kits are readily available in many parts of the country.
Zone 1 Predicted average indoor radon screening level greater than 4 pCi/L Zone 2 Predicted average indoor radon screening level between 2 and 4 pCi/L Zone 3 Predicted average indoor radon screening level less than 2 pCi/L 쑿 FIGURE 21.26 EPA map of radon zones in the United States. The color coding shows average indoor radon levels as a function of geographic location.
Radiation in the Environment and Living Systems
SECTION 21.9
907
CHEMISTRY AND LIFE RADIATION THERAPY Healthy cells are either destroyed or damaged by highenergy radiation, leading to physiological disorders. This radiation can also destroy unhealthy cells, however, including cancerous cells. All cancers are characterized by runaway cell growth that can produce malignant tumors. These tumors can be caused by the exposure of healthy cells to high-energy radiation. Paradoxically, however, they can be destroyed by the same radiation that caused them because the rapidly reproducing cells of the tumors are very susceptible to radiation damage. Thus, cancerous cells are more susceptible to destruction by radiation than healthy ones, allowing radiation to be used effectively in the treatment of cancer. As early as 1904, physicians used the radiation emitted by radioactive substances to treat tumors by destroying the mass of unhealthy tissue. The treatment of disease by high-energy radiation is called radiation therapy. Many radionuclides are currently used in radiation therapy. Most of them have short half-lives, meaning that they emit a great deal of radiation in a short period of time (씰 TABLE 21.10). The radiation source used in radiation therapy may be inside or outside the body. In almost all cases, radiation therapy uses gamma radiation emitted by radioisotopes. Any alpha or beta radiation that is emitted concurrently can be blocked by appropriate packaging. For example, 192Ir is often administered as “seeds” consisting of a core of radioactive isotope coated with 0.1 mm of platinum metal. The platinum coating stops the alpha and beta rays, but the gamma rays penetrate it readily. The radioactive seeds can be surgically implanted in a tumor.
SAMPLE INTEGRATIVE EXERCISE
In some cases, human physiology allows a radioisotope to be ingested. For example, most of the iodine in the human body ends up in the thyroid gland, so thyroid cancer can be treated by using large doses of 131I. Radiation therapy on deep organs, where a surgical implant is impractical, often uses a 60Co “gun” outside the body to shoot a beam of gamma rays at the tumor. Particle accelerators are also used as an external source of high-energy radiation for radiation therapy. Because gamma radiation is so strongly penetrating, it is nearly impossible to avoid damaging healthy cells during radiation therapy. Many cancer patients undergoing radiation treatment experience unpleasant and dangerous side effects such as fatigue, nausea, hair loss, a weakened immune system, and occasionally even death. However, if other treatments such as chemotherapy (the use of drugs to combat cancer) fail, radiation therapy can be a good option.
TABLE 21.10 • Some Radioisotopes Used in Radiation Therapy Isotope
Half-Life
Isotope
Half-Life
32
P Co 90 Sr 125 I
14.3 days 5.27 yr 28.8 yr 60.25 days
137
Cs Ir 198 Au 222 Rn
30 yr 74.2 days 2.7 days 3.82 days
131
8.04 days
226
1600 yr
60
I
Putting Concepts Together
Potassium ion is present in foods and is an essential nutrient in the human body. One of the naturally occurring isotopes of potassium, potassium-40, is radioactive. Potassium-40 has a natural abundance of 0.0117% and a half-life t1>2 = 1.28 * 109 yr. It undergoes radioactive decay in three ways: 98.2% is by electron capture, 1.35% is by beta emission, and 0.49% is by positron emission. (a) Why should we expect 40K to be radioactive? (b) Write the nuclear equations for the three modes by which 40K decays. (c) How many 40K + ions are present in 1.00 g of KCl? (d) How long does it take for 1.00% of the 40K in a sample to undergo radioactive decay? SOLUTION (a) The 40K nucleus contains 19 protons and 21 neutrons. There are very few stable nuclei with odd numbers of both protons and neutrons (Section 21.2). (b) Electron capture is capture of an inner-shell electron by the nucleus: + -10e 0 (-1e) by
40 19K
Beta emission is loss of a beta particle
40 19K
¡
¡
40 18Ar
the nucleus: 40 20Ca
+
0 -1e
Positron emission is loss of a positron (01e) by the nucleus: 0 ¡ 40 18Ar + 1e (c) The total number of K ions in the sample is +
(1.00 g KCl)a
40 19K
1 mol KCl 1 mol K + 6.022 * 1023 K + ba ba b = 8.08 * 1021 K + ions 74.55 g KCl 1 mol KCl 1 mol K +
192
Ra
908
CHAPTER 21
Nuclear Chemistry
Of these, 0.0117% are 40K + ions: (8.08 * 1021 K + ions)a
0.0117
40
K + ions
b = 9.45 * 1017 potassium-40 ions 100 ions (d) The decay constant (the rate constant) for the radioactive decay can be calculated from the half-life, using Equation 21.20: k =
+
0.693 0.693 = (5.41 * 10-10)>yr = t1>2 1.28 * 109 yr
The rate equation, Equation 21.19, then allows us to calculate the time required: ln ln
Nt = -kt N0
99 = -[(5.41 * 10-10)>yr]t 100
- 0.01005 = -[(5.41 * 10-10)>yr]t t =
-0.01005 = 1.86 * 107 yr (-5.41 * 10-10)>yr
That is, it would take 18.6 million years for just 1.00% of the 40K in a sample to decay.
CHAPTER SUMMARY AND KEY TERMS The nucleus of an atom contains protons and neutrons, both of which are called nucleons. Reactions that involve changes in atomic nuclei are called nuclear reactions. Nuclei that spontaneously change by emitting radiation are said to be radioactive. Radioactive nuclei are called radionuclides, and the atoms containing them are called radioisotopes. Radionuclides spontaneously change through a process called radioactive decay. The three most important types of radiation given off as a result of radioactive decay are alpha (A) particles (42 He), beta ( B ) particles (-10 e) , and gamma (G) radiation (00g). Positrons (01e), which are particles with the same mass as an electron but the opposite charge, can also be produced when a radioisotope decays. In nuclear equations, reactant and product nuclei are represented by giving their mass numbers and atomic numbers, as well as their chemical symbol. The totals of the mass numbers on both sides of the equation are equal; the totals of the atomic numbers on both sides are also equal. There are four common modes of radioactive decay: alpha decay, which reduces the atomic number by 2 and the mass number by 4, beta emission, which increases the atomic number by 1 and leaves the mass number unchanged, positron emission and electron capture, both of which reduce the atomic number by 1 and leave the mass number unchanged. INTRODUCTION AND SECTION 21.1
The neutron-to-proton ratio is an important factor determining nuclear stability. By comparing a nuclide’s neutron-toproton ratio with those in the band of stability, we can predict the mode of radioactive decay. In general, neutron-rich nuclei tend to emit beta particles; proton-rich nuclei tend to either emit positrons or undergo electron capture; and heavy nuclei tend to emit alpha particles. The presence of magic numbers of nucleons and an even number of protons and neutrons also help determine the stability of a nucleus. A nuclide may undergo a series of decay steps before a stable nuclide forms. This series of steps is called a radioactive series or a nuclear disintegration series. SECTION 21.2
SECTION 21.3 Nuclear transmutations, induced conversions of one nucleus into another, can be brought about by bombarding nuclei with either charged particles or neutrons. Particle accelerators increase the kinetic energies of positively charged particles, allowing these particles to overcome their electrostatic repulsion by the nucleus. Nuclear transmutations are used to produce the transuranium elements, those elements with atomic numbers greater than that of uranium.
The SI unit for the activity of a radioactive source is the becquerel (Bq), defined as one nuclear disintegration per second. A related unit, the curie (Ci), corresponds to 3.7 * 1010 disintegrations per second. Nuclear decay is a first-order process. The decay rate (activity) is therefore proportional to the number of radioactive nuclei. The half-life of a radionuclide, which is a constant, is the time needed for one-half of the nuclei to decay. Some radioisotopes can be used to date objects; 14C, for example, is used to date organic objects. Geiger counters and scintillation counters count the emissions from radioactive samples. The ease of detection of radioisotopes also permits their use as radiotracers to follow elements through reactions. SECTIONS 21.4 AND 21.5
SECTION 21.6 The energy produced in nuclear reactions is accompanied by measurable changes of mass in accordance with Einstein’s relationship, ¢E = c 2 ¢m. The difference in mass between nuclei and the nucleons of which they are composed is known as the mass defect. The mass defect of a nuclide makes it possible to calculate its nuclear binding energy, the energy required to separate the nucleus into individual nucleons. Energy is produced when heavy nuclei split (fission) and when light nuclei fuse (fusion).
Uranium-235, uranium-233, and plutonium-239 undergo fission when they capture a neutron, splitting into lighter nuclei and releasing more neutrons. The neutrons produced in one fission can cause further fission reactions, which can lead to a nuclear chain reaction. A reaction that maintains a constant rate
SECTIONS 21.7 AND 21.8
Exercises is said to be critical, and the mass necessary to maintain this constant rate is called a critical mass. A mass in excess of the critical mass is termed a supercritical mass. In nuclear reactors the fission rate is controlled to generate a constant power. The reactor core consists of fuel elements containing fissionable nuclei, control rods, a moderator, and a primary coolant. A nuclear power plant resembles a conventional power plant except that the reactor core replaces the fuel burner. There is concern about the disposal of highly radioactive nuclear wastes that are generated in nuclear power plants. Nuclear fusion requires high temperatures because nuclei must have large kinetic energies to overcome their mutual repulsions. Fusion reactions are therefore called thermonuclear reactions. It is not yet possible to generate power on Earth through a controlled fusion process.
909
Ionizing radiation is energetic enough to remove an electron from a water molecule; radiation with less energy is called nonionizing radiation. Ionizing radiation generates free radicals, reactive substances with one or more unpaired electrons. The effects of long-term exposure to low levels of radiation are not completely understood, but it is usually assumed that the extent of biological damage varies in direct proportion to the level of exposure. The amount of energy deposited in biological tissue by radiation is called the radiation dose and is measured in units of gray or rad. One gray (Gy) corresponds to a dose of 1 J>kg of tissue. The rad is a smaller unit; 100 rad = 1 Gy. The effective dose, which measures the biological damage created by the deposited energy, is measured in units of rem or sievert (Sv). The rem is obtained by multiplying the number of rad by the relative biological effectiveness (RBE); 100 rem = 1 Sv.
SECTION 21.9
KEY SKILLS • Write balanced nuclear equations. (Section 21.1) • Predict nuclear stability and expected type of nuclear decay from the neutron-to-proton ratio of an isotope. (Section 21.2) • Write balanced nuclear equations for nuclear transmutations. (Section 21.3) • Calculate ages of objects and/or the amount of a radionuclide remaining after a given period of time using the half-life of the radionuclide in question. (Section 21.4) • Calculate mass and energy changes for nuclear reactions. (Section 21.6) • Calculate the binding energies for nuclei. (Section 21.6) • Describe the difference between fission and fusion. (Sections 21.7 and 21.8) • Understand how a nuclear power plant operates and know the differences among various types of nuclear power plants. (Section 21.7) • Understand the meaning of radiation dosage terms. (Section 21.9) • Understand the biological effects of different kinds of radiation. (Section 21.9)
KEY EQUATIONS Nt = -kt N0 0.693 • k = t1>2 • ln
• E = mc 2
[21.19]
First-order rate law for nuclear decay
[21.20]
Relationship between nuclear decay constant and half-life; this is derived from the previous equation at Nt = 12N0
[21.22]
Einstein’s equation that relates mass and energy
EXERCISES VISUALIZING CONCEPTS
21.2 Write the balanced nuclear equation for the reaction represented by the diagram shown here. [Section 21.2]
64 Number of neutrons
21.1 Indicate whether each of the following nuclides lies within the belt of stability in Figure 21.2: (a) neon-24, (b) chlorine-32, (c) tin-108, (d) polonium-216. For any that do not, describe a nuclear decay process that would alter the neutron-to-proton ratio in the direction of increased stability. [Section 21.2]
63
62
61 45
46 47 Number of protons
48
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Nuclear Chemistry
CHAPTER 21
of its original concentration after 1 hour? [Sections 21.2, 21.4, and 21.5]
21.3 Draw a diagram similar to that shown in Exercise 21.2 that il4 207 lustrates the nuclear reaction 211 83 Bi ¡ 2He + 81 Tl. [Section 21.2] 21.4 The accompanying graph illustrates the decay of 88 42Mo, which decays via positron emission. (a) What is the half-life of the decay? (b) What is the rate constant for the decay? (c) What fraction of the original sample of 88 42Mo remains after 12 min? (d) What is the product of the decay process? [Section 21.4]
0.9
88 42Mo(g)
0.7
Mass of
0.8
0.5
0.6
110 min
9 Number of neutrons
1.0
10
2 min
7
10 min
6
0.4
64 s
5715 yr
8
71 s
0.3 0.2
20 min
5
0.1 5 0
2
4
10 6 8 Time (minutes)
12
14
16
21.5 All the stable isotopes of boron, carbon, nitrogen, oxygen, and fluorine are shown in the chart in the right hand column (in red), along with their radioactive isotopes with t1>2 7 1 min (in blue). (a) Write the chemical symbols, including mass and atomic numbers, for all of the stable isotopes. (b) Which radioactive isotopes are most likely to decay by beta emission? (c) Some of the isotopes shown are used in positron emission tomography. Which ones would you expect to be most useful for this application? (d) Which isotope would decay to 12.5%
6 7 8 Number of protons
9
21.6 The diagram shown here illustrates a fission process. (a) What is the unidentified product of the fission? (b) Use Figure 21.2 to predict whether the nuclear products of this fission reaction are stable. [Section 21.7]
239 94Pu
1 0n
95 40Zr 1 0n
⫹ 1 0n
?
RADIOACTIVITY (section 21.1) 21.7 Indicate the number of protons and neutrons in the following 201 nuclei: (a) 55 Hg, (c) potassium-39. 22Mn, (b) 21.8 Indicate the number of protons and neutrons in the following 37 nuclei: (a) 124 52 Te, (b) Cl, (c) thorium-232. 21.9 Give the symbol for (a) a neutron, (b) an alpha particle, (c) gamma radiation. 21.10 Give the symbol for (a) a proton, (b) a beta particle, (c) a positron. 21.11 Write balanced nuclear equations for the following processes: (a) rubidium-90 undergoes beta emission; (b) selenium-72 undergoes electron capture; (c) krypton-76 undergoes positron emission; (d) radium-226 emits alpha radiation. 21.12 Write balanced nuclear equations for the following transformations: (a) bismuth-213 undergoes alpha decay; (b) nitrogen-13 undergoes electron capture; (c) technicium-98 undergoes electron capture; (d) gold-188 decays by positron emission.
21.13 Decay of which nucleus will lead to the following products: (a) bismuth-211 by beta decay; (b) chromium-50 by positron emission; (c) tantalum-179 by electron capture; (d) radium226 by alpha decay? 21.14 What particle is produced during the following decay processes: (a) sodium-24 decays to magnesium-24; (b) mercury188 decays to gold-188; (c) iodine-122 decays to xenon-122; (d) plutonium-242 decays to uranium-238? 21.15 The naturally occurring radioactive decay series that begins 207 with 235 92 U stops with formation of the stable 82 Pb nucleus. The decays proceed through a series of alpha-particle and beta-particle emissions. How many of each type of emission are involved in this series? 21.16 A radioactive decay series that begins with 232 90 Th ends with formation of the stable nuclide 208 . How many alphaPb 82 particle emissions and how many beta-particle emissions are involved in the sequence of radioactive decays?
911
Exercises
NUCLEAR STABILITY (section 21.2)
21.18 Each of the following nuclei undergoes either beta decay or positron emission. Predict the type of emission for each: (a) tritium, 31H, (b) 89 38Sr, (c) iodine-120, (d) silver-102. 21.19 One of the nuclides in each of the following pairs is radioactive. 40 Predict which is radioactive and which is stable: (a) 39 19 K and 19 K, 209 208 (b) Bi and Bi, (c) nickel-58 and nickel-65. Explain. 21.20 One nuclide in each of these pairs is radioactive. Predict which 45 12 is radioactive and which is stable: (a) 40 20Ca and 20Ca, (b) C 14 and C, (c) lead-206 and thorium-230. Explain your choice in each case. 21.21 Which of the following nuclides have magic numbers of both protons and neutrons: (a) helium-4, (b) oxygen-18, (c) calcium40, (d) zinc-66, (e) lead-208? 21.22 Despite the similarities in the chemical reactivity of elements in the lanthanide series, their abundances in Earth’s crust vary by two orders of magnitude. This graph shows the relative abundance as a function of atomic number. How do you explain the sawtooth variation across the series?
5 Ce Log of abundance in Earth’s crust (parts per billion by mass)
21.17 Predict the type of radioactive decay process for the following radionuclides: (a) 85B, (b) 68 29Cu , (c) phosphorus-32, (d) chlorine-39.
Nd La 4 Gd
Sm
Pr
Dy Er
Yb
Eu 3
Tb
Ho Lu
Tm
2
56
58
60
62 64 66 Atomic number
68
70
72
21.23 Using the concept of magic numbers, explain why alpha emission is relatively common, but proton emission is nonexistent. 21.24 Which of the following nuclides would you expect to be ra58 108 dioactive: 62 28Ni, 29Cu, 47 Ag, tungsten-184, polonium-206? Justify your choices.
NUCLEAR TRANSMUTATIONS (section 21.3) 21.25 Why are nuclear transmutations involving neutrons generally easier to accomplish than those involving protons or alpha particles? 21.26 In 1930 the American physicist Ernest Lawrence designed the first cyclotron in Berkeley, California. In 1937 Lawrence bombarded a molybdenum target with deuterium ions, producing for the first time an element not found in nature. What was this element? Starting with molybdenum-96 as your reactant, write a nuclear equation to represent this process. 21.27 Complete and balance the following nuclear equations by supplying the missing particle: 10 1 (a) 252 98 Cf + 5 B ¡ 3 0 n + ? 2 3 4 (b) 1 H + 2 He ¡ 2 He + ? (c) 11 H + 115 B ¡ 3?
(d) (e)
122 122 53 I ¡ 54 Xe + 59 0 26 Fe ¡ -1 e + ?
?
21.28 Complete and balance the following nuclear equations by supplying the missing particle: (a) 147 N + 42 He ¡ ? + 11 H 0 (b) 40 19 K + -1 e (orbital electron) ¡ ? 4 1 (c) ? + 2 He ¡ 30 14 Si + 1 H 1 60 (d) 58 26 Fe + 2 0 n ¡ 27 Co + ? 235 1 135 (e) 92 U + 0 n ¡ 54 Xe + 2 10 n + ? 241 21.29 Write balanced equations for (a) 238 92 U(a, n) 94 Pu, 60 (b) 147N(a, p)178O, (c) 56 . Fe(a, b) Cu 26 29
21.30 Write balanced equations for each of the following nuclear reac239 14 11 18 19 tions: (a) 238 92 U(n, g) 92 U, (b) 7 N(p, a) 6 C, (c) 8 O(n, b) 9 F.
RATES OF RADIOACTIVE DECAY (section 21.4) 21.31 Each statement that follows refers to a comparison between two radioisotopes, A and X. Indicate whether each of the following statements is true or false, and why. (a) If the half-life for A is shorter than the half-life for X, A has a larger decay rate constant. (b) If X is “not radioactive,” its half-life is essentially zero. (c) If A has a half-life of 10 years, and X has a half-life of 10,000 years, A would be a more suitable radioisotope to measure processes occurring on the 40-year time scale. 21.32 It has been suggested that strontium-90 (generated by nuclear testing) deposited in the hot desert will undergo radioactive decay more rapidly because it will be exposed to much higher
average temperatures. (a) Is this a reasonable suggestion? (b) Does the process of radioactive decay have an activation energy, like the Arrhenius behavior of many chemical reactions (Section 14.5)? Discuss. 21.33 Some watch dials are coated with a phosphor, like ZnS, and a polymer in which some of the 1H atoms have been replaced by 3 H atoms, tritium. The phosphor emits light when struck by the beta particle from the tritium decay, causing the dials to glow in the dark. The half-life of tritium is 12.3 yr. If the light given off is assumed to be directly proportional to the amount of tritium, by how much will a dial be dimmed in a watch that is 50 years old?
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CHAPTER 21
Nuclear Chemistry
21.34 It takes 5.2 min for a 1.000-g sample of to 0.250 g. What is the half-life of 210Fr?
210
Fr to decay
21.35 Cobalt-60 is a strong gamma emitter that has a half-life of 5.26 yr. The cobalt-60 in a radiotherapy unit must be replaced when its radioactivity falls to 75% of the original sample. If an original sample was purchased in June 2010, when will it be necessary to replace the cobalt-60? 21.36 How much time is required for a 6.25-mg sample of decay to 0.75 mg if it has a half-life of 27.8 days?
51
Cr to
[21.37] Radium-226, which undergoes alpha decay, has a half-life of 1600 yr. (a) How many alpha particles are emitted in 5.0 min by a 10.0-mg sample of 226Ra? (b) What is the activity of the sample in mCi? [21.38] Cobalt-60, which undergoes beta decay, has a half-life of 5.26 yr. (a) How many beta particles are emitted in 600 s by a 3.75-mg sample of 60Co? (b) What is the activity of the sample in Bq?
21.39 The cloth shroud from around a mummy is found to have a 14 C activity of 9.7 disintegrations per minute per gram of carbon as compared with living organisms that undergo 16.3 disintegrations per minute per gram of carbon. From the half-life for 14C decay, 5715 yr, calculate the age of the shroud. 21.40 A wooden artifact from a Chinese temple has a 14C activity of 38.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. From the halflife for 14C decay, 5715 yr, determine the age of the artifact. 21.41 Potassium-40 decays to argon-40 with a half-life of 1.27 * 109 yr. What is the age of a rock in which the mass ratio of 40Ar to 40K is 4.2? 21.42 The half-life for the process 238U ¡ 206Pb is 4.5 * 109 yr. A mineral sample contains 75.0 mg of 238U and 18.0 mg of 206 Pb. What is the age of the mineral?
ENERGY CHANGES (section 21.6) 21.43 The thermite reaction, Fe 2 O3(s) + 2 Al(s) ¡ 2 Fe(s) + Al 2 O3(s), ¢H° = -851.5 kJ/mol, is one of the most exothermic reactions known. Because the heat released is sufficient to melt the iron product, the reaction is used to weld metal under the ocean. How much heat is released per mole of Fe2O3 produced? How does this amount of thermal energy compare with the energy released when 2 mol of protons and 2 mol of neutrons combine to form 1 mol of alpha particles? 21.44 An analytical laboratory balance typically measures mass to the nearest 0.1 mg. What energy change would accompany the loss of 0.1 mg in mass? 21.45 How much energy must be supplied to break a single aluminum-27 nucleus into separated protons and neutrons if an aluminum-27 atom has a mass of 26.9815386 amu? How much energy is required for 100.0 grams of aluminum-27? (The mass of an electron is given on the inside back cover.) 21.46 How much energy must be supplied to break a single 21Ne nucleus into separated protons and neutrons if the nucleus has a mass of 20.98846 amu? What is the nuclear binding energy for 1 mol of 21Ne? 21.47 The atomic masses of hydrogen-2 (deuterium), helium-4, and lithium-6 are 2.014102 amu, 4.002602 amu, and 6.0151228 amu, respectively. For each isotope, calculate (a) the nuclear mass, (b) the nuclear binding energy, (c) the nuclear binding energy per nucleon. (d) Which of these three isotopes has the largest nuclear binding energy per nucleon? Does this agree with the trends plotted in Figure 21.12? 21.48 The atomic masses of nitrogen-14, titanium-48, and xenon129 are 13.999234 amu, 47.935878 amu, and 128.904779 amu, respectively. For each isotope, calculate (a) the nuclear mass,
(b) the nuclear binding energy, (c) the nuclear binding energy per nucleon. 21.49 The energy from solar radiation falling on Earth is 1.07 * 1016 kJ/min. (a) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (b) If the energy released in the reaction 235
U + 10 n ¡
141 56 Ba
+
92 36 Kr
+ 3 10 n
(235U nuclear mass, 234.9935 amu; 141Ba nuclear mass, 140.8833 amu; 92Kr nuclear mass, 91.9021 amu) is taken as typical of that occurring in a nuclear reactor, what mass of uranium-235 is required to equal 0.10% of the solar energy that falls on Earth in 1.0 day? 21.50 Based on the following atomic mass values—1H, 1.00782 amu; 2H, 2.01410 amu; 3H, 3.01605 amu; 3He, 3.01603 amu; 4He, 4.00260 amu—and the mass of the neutron given in the text, calculate the energy released per mole in each of the following nuclear reactions, all of which are possibilities for a controlled fusion process: (a) 21 H + 31 H ¡ 42 He + 10 n (b) 21 H + 21 H ¡ 32 He + 10 n (c) 21 H + 32 He ¡ 42 He + 11 H 21.51 Which of the following nuclei is likely to have the largest mass defect per nucleon: (a) 59Co, (b) 11B, (c) 118Sn, (d) 243Cm? Explain your answer. 21.52 The isotope 62 28 Ni has the largest binding energy per nucleon of any isotope. Calculate this value from the atomic mass of nickel-62 (61.928345 amu) and compare it with the value given for iron-56 in Table 21.7.
EFFECTS AND USES OF RADIOISOTOPES (sections 21.7–21.9) 21.53 Iodine-131 is a convenient radioisotope to monitor thyroid activity in humans. It is a beta emitter with a half-life of 8.02 days. The thyroid is the only gland in the body that uses iodine. A person undergoing a test of thyroid activity drinks a solution of NaI, in which only a small fraction of the iodide is radioactive. (a) Why is NaI a good choice for the source of iodine? (b) If a Geiger counter
is placed near the person’s thyroid (which is near the neck) right after the sodium iodide solution is taken, what will the data look like as a function of time? (c) A normal thyroid will take up about 12% of the ingested iodide in a few hours. How long will it take for the radioactive iodide taken up and held by the thyroid to decay to 0.01% of the original amount?
Additional Exercises 21.54 Why is it important that radioisotopes used as diagnostic tools in nuclear medicine produce gamma radiation when they decay? Why are alpha emitters not used as diagnostic tools? 21.55 What is the most common fissionable isotope in a commercial nuclear power reactor? 21.56 What is meant by enriched uranium? How is enriched uranium different from natural uranium? 21.57 What is the function of the control rods in a nuclear reactor? What substances are used to construct control rods? Why are these substances chosen? 21.58 (a) What is the function of the moderator in a nuclear reactor? (b) What substance acts as the moderator in a pressurized water generator? (c) What other substances are used as a moderator in nuclear reactor designs? 21.59 Complete and balance the nuclear equations for the following fission or fusion reactions: (a) 21H + 21H ¡ 32He + — 1 133 98 1 (b) 239 92 U + 0n ¡ 51 Sb + 41Nb + — 0n 21.60 Complete and balance the nuclear equations for the following fission reactions: 1 160 72 1 (a) 235 92 U + 0n ¡ 62 Sm + 30Zn + — 0n 239 1 144 1 (b) 94Pu + 0n ¡ 58Ce + — + 20n 21.61 A portion of the Sun’s energy comes from the reaction 4 11H ¡ 42He + 2 01e which requires a temperature of 106 to 107 K. (a) Use the mass of the helium-4 nucleus given in Table 21.7 to determine how much energy is released when the reaction is run with 1 mol of hydrogen atoms. (b) Why is such a high temperature required? 21.62 The spent fuel elements from a fission reactor are much more intensely radioactive than the original fuel elements. (a) What does this tell you about the products of the fission process in relationship to the belt of stability, Figure 21.2? (b) Given that only two or three neutrons are released per fission event and knowing that the nucleus undergoing fission has a neutronto-proton ratio characteristic of a heavy nucleus, what sorts of decay would you expect to be dominant among the fission products?
913
21.63 Which type or types of nuclear reactors have these characteristics? (a) Does not use a secondary coolant (b) Creates more fissionable material than it consumes (c) Uses a gas, such as He or CO2, as the primary coolant 21.64 Which type or types of nuclear reactors have these characteristics? (a) Can use natural uranium as a fuel (b) Does not use a moderator (s) Can be refueled without shutting down 21.65 Hydroxyl radicals can pluck hydrogen atoms from molecules (“hydrogen abstraction”), and hydroxide ions can pluck protons from molecules (“deprotonation”). Write the reaction equations and Lewis dot structures for the hydrogen abstraction and deprotonation reactions for the generic carboxylic acid R–COOH with hydroxyl radical and hydroxide ion, respectively. Why is hydroxyl radical more toxic to living systems than hydroxide ion? 21.66 Which are classified as ionizing radiation: X-rays, alpha particles, microwaves from a cell phone, and gamma rays? 21.67 A laboratory rat is exposed to an alpha-radiation source whose activity is 14.3 mCi. (a) What is the activity of the radiation in disintegrations per second? In becquerels? (b) The rat has a mass of 385 g and is exposed to the radiation for 14.0 s, absorbing 35% of the emitted alpha particles, each having an energy of 9.12 * 10-13 J. Calculate the absorbed dose in millirads and grays. (c) If the RBE of the radiation is 9.5, calculate the effective absorbed dose in mrem and Sv. 21.68 A 65-kg person is accidentally exposed for 240 s to a 15-mCi source of beta radiation coming from a sample of 90Sr. (a) What is the activity of the radiation source in disintegrations per second? In becquerels? (b) Each beta particle has an energy of 8.75 * 10-14 J, and 7.5% of the radiation is absorbed by the person. Assuming that the absorbed radiation is spread over the person’s entire body, calculate the absorbed dose in rads and in grays. (c) If the RBE of the beta particles is 1.0, what is the effective dose in mrem and in sieverts? (d) How does the magnitude of this dose of radiation compare with that of a mammogram (300 mrem)?
ADDITIONAL EXERCISES 21.69 Radon-222 decays to a stable nucleus by a series of three alpha emissions and two beta emissions. What is the stable nucleus that is formed? 21.70 Equation 21.28 is the nuclear reaction responsible for much of the helium-4 production in our Sun. How much energy is released in this reaction? 21.71 Chlorine has two stable nuclides, 35Cl and 37Cl. In contrast, 36 Cl is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of 36Cl? (b) Based on the empirical rules about nuclear stability, explain why the nucleus of 36Cl is less stable than either 35Cl or 37Cl. 21.72 When two protons fuse in a star, the product is 2H plus a positron (Equation 21.26). Why do you think the more obvious product of the reaction, 2He, is unstable? 21.73 Nuclear scientists have synthesized approximately 1600 nuclei not known in nature. More might be discovered with heavyion bombardment using high-energy particle accelerators. Complete and balance the following reactions, which involve heavy-ion bombardments:
(a) 63Li + 56 28Ni ¡ ? 40 147 (b) 20Ca + 248 96 Cm ¡ 62 Sm + ? 88 84 116 (c) 38Sr + 36Kr ¡ 46Pd + ? 238 70 1 (d) 40 20Ca + 92 U ¡ 30Zn + 4 0n + 2? [21.74] The synthetic radioisotope technetium-99, which decays by beta emission, is the most widely used isotope in nuclear medicine. The following data were collected on a sample of 99Tc: Disintegrations per Minute
Time (h)
180 130 104 77 59 46 24
0 2.5 5.0 7.5 10.0 12.5 17.5
Using these data, make an appropriate graph and curve fit to determine the half-life.
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CHAPTER 21
Nuclear Chemistry
[21.75] According to current regulations, the maximum permissible dose of strontium-90 in the body of an adult is 1 mCi (1 * 10-6 Ci). Using the relationship rate = kN, calculate the number of atoms of strontium-90 to which this dose corresponds. To what mass of strontium-90 does this correspond? The half-life for strontium-90 is 28.8 yr. [21.76] Suppose you had a detection device that could count every decay event from a radioactive sample of plutonium-239 (t1>2 is 24,000 yr). How many counts per second would you obtain from a sample containing 0.385 g of plutonium-239? 21.77 Methyl acetate (CH3COOCH3) is formed by the reaction of acetic acid with methyl alcohol. If the methyl alcohol is labeled with oxygen-18, the oxygen-18 ends up in the methyl acetate: O CH3COH ⫹ H18OCH3
O CH3C18OCH3 ⫹ H2O
Do the C ¬ OH bond of the acid and the O ¬ H bond of the alcohol break in the reaction, or do the O ¬ H bond of the acid and the C ¬ OH bond of the alcohol break? Explain. 21.78 An experiment was designed to determine whether an aquatic plant absorbed iodide ion from water. Iodine-131 (t1>2 = 8.02 days) was added as a tracer, in the form of iodide ion, to a tank containing the plants. The initial activity of a 1.00-mL sample of the water was 214 counts per minute. After 30 days the level of activity in a 1.00-mL sample was 15.7 counts per minute. Did the plants absorb iodide from the water? Explain.
21.79 The nuclear masses of 7Be, 9Be, and 10Be are 7.0147, 9.0100, and 10.0113 amu, respectively. Which of these nuclei has the largest binding energy per nucleon? [21.80] A 26.00-g sample of water containing tritium, 31 H, emits 1.50 * 103 beta particles per second. Tritium is a weak beta emitter with a half-life of 12.3 yr. What fraction of all the hydrogen in the water sample is tritium? 21.81 The Sun radiates energy into space at the rate of 3.9 * 1026 J>s. (a) Calculate the rate of mass loss from the Sun in kg>s. (b) How does this mass loss arise? (c) It is estimated that the Sun contains 9 * 1056 free protons. How many protons per second are consumed in nuclear reactions in the Sun? [21.82] The average energy released in the fission of a single uranium-235 nucleus is about 3 * 10-11 J. If the conversion of this energy to electricity in a nuclear power plant is 40% efficient, what mass of uranium-235 undergoes fission in a year in a plant that produces 1000 megawatts? Recall that a watt is 1 J>s. 21.83 Tests on human subjects in Boston in 1965 and 1966, following the era of atomic bomb testing, revealed average quantities of about 2 pCi of plutonium radioactivity in the average person. How many disintegrations per second does this level of activity imply? If each alpha particle deposits 8 * 10-13 J of energy and if the average person weighs 75 kg, calculate the number of rads and rems of radiation in 1 yr from such a level of plutonium.
Integrative Exercises
915
INTEGRATIVE EXERCISES 21.84 A 53.8-mg sample of sodium perchlorate contains radioactive chlorine-36 (whose atomic mass is 36.0 amu). If 29.6% of the chlorine atoms in the sample are chlorine-36 and the remainder are naturally occurring nonradioactive chlorine atoms, how many disintegrations per second are produced by this sample? The half-life of chlorine-36 is 3.0 * 105 yr. 21.85 Calculate the mass of octane, C8H18(l), that must be burned in air to evolve the same quantity of energy as produced by the fusion of 1.0 g of hydrogen in the following fusion reaction: 4 11 H ¡ 42 He + 2 01 e Assume that all the products of the combustion of octane are in their gas phases. Use data from Exercise 21.50, Appendix C, and the inside covers of the text. The standard enthalpy of formation of octane is -250.1 kJ/mol. 21.86 A sample of an alpha emitter having an activity of 0.18i is stored in a 25.0-mL sealed container at 22 °C for 245 days. (a) How many alpha particles are formed during this time? (b) Assuming that each alpha particle is converted to a helium atom, what is the partial pressure of helium gas in the container after this 245-day period?
[21.87] Charcoal samples from Stonehenge in England were burned in O2, and the resultant CO2 gas bubbled into a solution of Ca(OH)2 (limewater), resulting in the precipitation of CaCO3. The CaCO3 was removed by filtration and dried. A 788-mg sample of the CaCO3 had a radioactivity of 1.5 * 10-2 Bq due to carbon-14. By comparison, living organisms undergo 15.3 disintegrations per minute per gram of carbon. Using the half-life of carbon-14, 5715 yr, calculate the age of the charcoal sample. 21.88 A 25.0-mL sample of 0.050 M barium nitrate solution was mixed with 25.0 mL of 0.050 M sodium sulfate solution labeled with radioactive sulfur-35. The activity of the initial sodium sulfate solution was 1.22 * 106 Bq>mL. After the resultant precipitate was removed by filtration, the remaining filtrate was found to have an activity of 250 Bq>mL. (a) Write a balanced chemical equation for the reaction that occurred. (b) Calculate the Ksp for the precipitate under the conditions of the experiment.
WHAT’S AHEAD 22.1 PERIODIC TRENDS AND CHEMICAL REACTIONS
22.4 GROUP 7A: THE HALOGENS
We begin with a review of periodic trends and types of chemical reactions, which will help us focus on general patterns of behavior as we examine each family in the periodic table.
We then explore the most electronegative elements: the halogens, group 7A.
22.5 OXYGEN 22.2 HYDROGEN The first nonmetal we consider, hydrogen, forms compounds with most other nonmetals and with many metals.
We next consider oxygen, the most abundant element by mass in both Earth’s crust and the human body, and the oxide and peroxide compounds it forms.
22.3 GROUP 8A: THE NOBLE GASES
22.6 THE OTHER GROUP 6A ELEMENTS: S, Se, Te, AND Po
Next, we consider the noble gases, the elements of group 8A, which exhibit very limited chemical reactivity.
We study the other members of group 6A (S, Se, Te, and Po), of which sulfur is the most important.
22
AN OCEAN SUNSET SEEN THROUGH a glass window.
22.7 NITROGEN
22.9 CARBON
We next consider nitrogen, a key component of our atmosphere. It forms compounds in which its oxidation number ranges from - 3 to +5, including such important compounds as NH3 and HNO3.
We next focus on the inorganic compounds of carbon.
22.8 THE OTHER GROUP 5A ELEMENTS: P, As, Sb, AND Bi Of the other members of group 5A (P, As, Sb, and Bi), we take a closer look at phosphorus—the most commercially important one and the only one that plays an important and beneficial role in biological systems.
22.10 THE OTHER GROUP 4A ELEMENTS: Si, Ge, Sn, AND Pb We then consider silicon, the element most abundant and significant of the heavier members of group 4A.
22.11 BORON Finally, we examine boron—the sole nonmetallic element of group 3A.
CHEMISTRY OF THE NONMETALS shows the sun setting over the ocean, viewed through a window. Everything we see there comes from nonmetallic elements. The heat and light of the Sun result from the nuclear reactions of hydrogen. We know that water is H2O; glass, too, is nonmetallic, being based on silicon dioxide, SiO2.
THE CHAPTER-OPENING PHOTOGRAPH
In this chapter we take a panoramic view of the descriptive chemistry of the nonmetallic elements, starting with hydrogen and progressing group by group across the periodic table. We will consider how the elements occur in nature, how they are isolated from their sources, and how they are used. We will emphasize hydrogen, oxygen, nitrogen, and carbon because these four nonmetals form many commercially important compounds and account for 99% of the atoms required by living cells. As you study this descriptive chemistry, it is important to look for trends rather than trying to memorize all the facts presented. The periodic table is your most valuable tool in this task.
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CHAPTER 22
Chemistry of the Nonmetals
|
22.1 PERIODIC TRENDS AND CHEMICAL REACTIONS Metals
Metalloids
Recall that we can classify elements as metals, metalloids, and nonmetals. • (Section 7.6) Except for hydrogen, which is a special case, the nonmetals occupy the upper right portion of the periodic table. This division of elements relates nicely to trends in the properties of the elements as summarized in 씱 FIGURE 22.1. Electronegativity, for example, increases as we move left to right across a period and decreases as we move down a group. The nonmetals thus have higher electronegativities than the metals. This difference leads to the formation of ionic solids in reactions between metals and nonmetals. • (Sections 7.6, 8.2, 8.4) In contrast, compounds formed between two or more nonmetals are usually molecular substances. • (Sections 7.8 and 8.4) The chemistry exhibited by the first member of a nonmetal group can differ from that of subsequent members. For example, nonmetals in period 3 and below can accommodate a larger number of bonded neighbors. • (Section 8.7) Another important difference is that the first element in any group can more readily form p bonds. This trend is due, in part, to size because small atoms are able to approach each other more closely. As a result, the overlap of p orbitals, which results in the formation of p bonds, is more effective for the first element in each group (쑼 FIGURE 22.2). More effective overlap means stronger p bonds, reflected in bond enthalpies. • (Section 8.8) For example, the difference between the enthalpies of the C ¬ C bond and the C “ C bond is about 270 kJ> mol; • (Table 8.4) this value reflects the “strength” of a carbon–carbon p bond, and the difference between Si ¬ Si and Si “ Si bonds is only about 100 kJ> mol, significantly lower than that for carbon. As we shall see, p bonds are particularly important in the chemistry of carbon, nitrogen, and oxygen, each the first member in its group. The heavier elements in these groups have a tendency to form only single bonds.
Nonmetals
Decreasing ionization energy Increasing atomic radius Decreasing electronegativity Increasing metallic character
Increasing ionization energy Decreasing atomic radius Increasing electronegativity Decreasing metallic character
쑿 FIGURE 22.1 Trends in elemental properties.
SAMPLE EXERCISE 22.1
Identifying Elemental Properties
Of the elements Li, K, N, P, and Ne, which (a) is the most electronegative, (b) has the greatest metallic character, (c) can bond to more than four atoms in a molecule, (d) forms p bonds most readily? SOLUTION Analyze We are given a list of elements and asked to predict several properties that can be related to periodic trends. Plan We can use Figures 22.1 and 22.2 to guide us to the answers.
C C
Smaller nucleusto-nucleus distance, more orbital overlap, stronger bond
Si
Si
Larger nucleusto-nucleus distance, less orbital overlap, weaker bond
쑿 FIGURE 22.2 P bonds in period 2 and period 3 elements.
Solve (a) Electronegativity increases as we proceed toward the upper right portion of the periodic table, excluding the noble gases. Thus, N is the most electronegative element of our choices. (b) Metallic character correlates inversely with electronegativity—the less electronegative an element, the greater its metallic character. The element with the greatest metallic character is therefore K, which is closest to the lower left corner of the periodic table. (c) Nonmetals tend to form molecular compounds, so we can narrow our choice to the three nonmetals on the list: N, P, and Ne. To form more than four bonds, an element must be able to expand its valence shell to allow more than an octet of electrons around it. Valence-shell expansion occurs for period 3 elements and below; N and Ne are both in period 2 and do not undergo valence-shell expansion. Thus, the answer is P. (d) Period 2 nonmetals form p bonds more readily than elements in period 3 and below. There are no compounds known that contain covalent bonds to Ne. Thus, N is the element from the list that forms p bonds most readily.
SECTION 22.1
Periodic Trends and Chemical Reactions
919
PRACTICE EXERCISE Of the elements Be, C, Cl, Sb, and Cs, which (a) has the lowest electronegativity, (b) has the greatest nonmetallic character, (c) is most likely to participate in extensive p bonding, (d) is most likely to be a metalloid? Answers: (a) Cs, (b) Cl, (c) C, (d) Sb
The ready ability of period 2 elements to form p bonds is an important factor in determining the structures of these elements and their compounds. Compare, for example, the elemental forms of carbon and silicon. Carbon has five major crystalline allotropes: diamond, graphite, buckminsterfullerene, graphene, and carbon nanotubes. • (Sections 12.7, 12.9) Diamond is a covalent-network solid that has C ¬ C s bonds but no p bonds. Graphite, buckminsterfullerene, graphene, and carbon nanotubes have p bonds that result from the sideways overlap of p orbitals. Elemental silicon, however, exists only as a diamondlike covalent-network solid with s bonds; it has no forms analogous to graphite, buckminsterfullerene, graphene, or carbon nanotubes, apparently because Si ¬ Si p bonds are weak. GIVE IT SOME THOUGHT Can silicon–silicon double bonds form in elemental silicon?
We likewise see significant differences in the dioxides of carbon and silicon (씰 FIGURE 22.3). CO2 is a molecular substance containing C “ O double bonds, whereas SiO2 is a covalent-network solid in which four oxygen atoms are bonded to each silicon atom by single bonds, forming an extended structure that has the empirical formula SiO2. GIVE IT SOME THOUGHT Nitrogen is found in nature as N2(g). Would you expect phosphorus to be found in nature as P2(g)? Explain.
Fragment of extended SiO2 lattice; Si forms only single bonds
Chemical Reactions Because O2 and H2O are abundant in our environment, it is particularly important to consider how these substances react with other compounds. About one-third of the reactions discussed in this chapter involve either O2 (oxidation or combustion reactions) or H2O (especially proton-transfer reactions). In combustion reactions, • (Section 3.2) hydrogen-containing compounds produce H2O. Carbon-containing ones produce CO2 (unless the amount of O2 is insufficient, in which case CO or even C can form). Nitrogen-containing compounds tend to form N2, although NO can form in special cases or in small amounts. A reaction illustrating these points is: 4 CH3NH2(g) + 9 O2(g) ¡ 4 CO2(g) + 10 H2O(g) + 2 N2(g)
[22.1]
The formation of H2O, CO2, and N2 reflects the high thermodynamic stability of these substances, indicated by the large bond energies for the O ¬ H, C “ O, and N ‚ N bonds (463, 799, and 941 kJ> mol, respectively). • (Section 8.8) When dealing with proton-transfer reactions, remember that the weaker a Brønsted–Lowry acid, the stronger its conjugate base. • (Section 16.2) For example, H2, OH-, NH3, and CH4 are exceedingly weak proton donors that have no tendency to act as acids in water. Thus, the species formed by removing one or more protons from them are extremely strong bases. All react readily with water, removing protons from H2O to form OH- . Two representative reactions are: CH3-(aq) + H2O(l) ¡ CH4(g) + OH-(aq)
[22.2]
N3-(aq) + 3 H2O(l) ¡ NH3(aq) + 3 OH-(aq)
[22.3]
CO2 ; C forms double bonds 쑿 FIGURE 22.3 Comparison of the bonds in SiO2 and CO2.
920
CHAPTER 22
Chemistry of the Nonmetals
SAMPLE EXERCISE 22.2
Predicting the Products of Chemical Reactions
Predict the products formed in each of the following reactions, and write a balanced equation: (a) CH3NHNH2(g) + O2(g) ¡ (b) Mg 3P2(s) + H2O(l) ¡ SOLUTION Analyze: We are given the reactants for two chemical equations and asked to predict the products and then balance the equations. Plan We need to examine the reactants to see if we might recognize a reaction type. In (a) the carbon compound is reacting with O2, which suggests a combustion reaction. In (b) water reacts with an ionic Solve (a) Based on the elemental composition of the carbon compound, this combustion reaction should produce CO2, H2O, and N2: 2+
compound. The anion, P3-, is a strong base and H2O is able to act as an acid, so the reactants suggest an acid–base (proton-transfer) reaction.
2 CH3NHNH2(g) + 5 O2(g) ¡ 2 CO2(g) + 6 H2O(g) + 2 N2(g)
3-
(b) Mg3P2 is ionic, consisting of Mg and P ions. The P3- ion, like N3- , has a strong affinity for protons and reacts with H2O to form OH- and PH3 (PH2- , PH2 - , and PH3 are all exceedingly weak proton donors).
Mg 3P2(s) + 6 H2O(l) ¡ 2 PH3(g) + 3 Mg(OH)2(s)
Mg(OH)2 has low solubility in water and will precipitate. PRACTICE EXERCISE Write a balanced equation for the reaction of solid sodium hydride with water. Answer: NaH(s) + H2O(l) ¡ NaOH(aq) + H2(g)
22.2 | HYDROGEN The English chemist Henry Cavendish (1731–1810) was the first to isolate hydrogen. Because the element produces water when burned in air, the French chemist Antoine Lavoisier • (Figure 3.1) gave it the name hydrogen, which means “water producer” (Greek: hydro, water; gennao, to produce). Hydrogen is the most abundant element in the universe. It is the nuclear fuel consumed by our Sun and other stars to produce energy. • (Section 21.8) Although about 75% of the known mass of the universe is hydrogen, it constitutes only 0.87% of Earth’s mass. Most of the hydrogen on our planet is found associated with oxygen. Water, which is 11% hydrogen by mass, is the most abundant hydrogen compound.
Isotopes of Hydrogen The most common isotope of hydrogen, 11H, has a nucleus consisting of a single proton. This isotope, sometimes referred to as protium,* makes up 99.9844% of naturally occurring hydrogen. Two other isotopes are known: 21H, whose nucleus contains a proton and a neutron, 3 and 1H, whose nucleus contains a proton and two neutrons. The 21H isotope, deuterium, makes up 0.0156% of naturally occurring hydrogen. It is not radioactive and is often given the symbol D in chemical formulas, as in D2O (deuterium oxide), which is known as heavy water. Because an atom of deuterium is about twice as massive as an atom of protium, the properties of deuterium-containing substances vary somewhat from those of the protiumcontaining analogs. For example, the normal melting and boiling points of D2O are 3.81 °C and 101.42 °C, respectively, versus 0.00 °C and 100.00 °C for H2O. Not surprisingly, the *Giving unique names to isotopes is limited to hydrogen. Because of the proportionally large differences in their masses, the isotopes of H show appreciably more differences in their properties than isotopes of heavier elements.
SECTION 22.2
density of D2O at 25 °C (1.104 g> mL) is greater than that of H2O (0.997 g> mL). Replacing protium with deuterium (a process called deuteration) can also have a profound effect on reaction rates, a phenomenon called a kinetic-isotope effect. For example, heavy water can be obtained from the electrolysis 32 H2O(l) ¡ 2 H2(g) + O2(g)4 of ordinary water because the small amount of naturally occurring D2O in the sample undergoes electrolysis more slowly than H2O and, therefore, becomes concentrated during the reaction. The third isotope, 31H, tritium, is radioactive, with a half-life of 12.3 yr: 3 1H
¡ 32He +
0 -1e
t1>2 = 12.3 yr
[22.4]
Because of its short half-life, only trace quantities of tritium exist naturally. The isotope can be synthesized in nuclear reactors by neutron bombardment of lithium-6: 6 3Li
+ 10n ¡ 31H + 42He
[22.5]
Deuterium and tritium are useful in studying reactions of compounds containing hydrogen. A compound is “labeled” by replacing one or more ordinary hydrogen atoms with deuterium or tritium at specific locations in a molecule. By comparing the locations of the label atoms in reactants and products, the reaction mechanism can often be inferred. When methyl alcohol (CH3OH) is placed in D2O, for example, the H atom of the O ¬ H bond exchanges rapidly with the D atoms, forming CH3OD. The H atoms of the CH3 group do not exchange. This experiment demonstrates the kinetic stability of C ¬ H bonds and reveals the speed at which the O ¬ H bond in the molecule breaks and re-forms.
Properties of Hydrogen Hydrogen is the only element that is not a member of any family in the periodic table. Because of its ls1 electron configuration, it is generally placed above lithium in the table. However, it is definitely not an alkali metal. It forms a positive ion much less readily than any alkali metal. The ionization energy of the hydrogen atom is 1312 kJ> mol, whereas that of lithium is 520 kJ> mol. Hydrogen is sometimes placed above the halogens in the periodic table because the hydrogen atom can pick up one electron to form the hydride ion, H-, which has the same electron configuration as helium. However, the electron affinity of hydrogen, E = - 73 kJ>mol, is not as large as that of any halogen. In general, hydrogen shows no closer resemblance to the halogens than it does to the alkali metals. Elemental hydrogen exists at room temperature as a colorless, odorless, tasteless gas composed of diatomic molecules. We can call H2 dihydrogen, but it is more commonly referred to as either molecular hydrogen or simply hydrogen. Because H2 is nonpolar and has only two electrons, attractive forces between molecules are extremely weak. As a result, its melting point (-259 °C) and boiling point ( -253 °C) are very low. The H ¬ H bond enthalpy (436 kJ> mol) is high for a single bond. • (Table 8.4) By comparison, the Cl ¬ Cl bond enthalpy is only 242 kJ> mol. Because H2 has a strong bond, most reactions involving H2 are slow at room temperature. However, the molecule is readily activated by heat, irradiation, or catalysis. The activation generally produces hydrogen atoms, which are very reactive. Once H2 is activated, it reacts rapidly and exothermically with a wide variety of substances. GIVE IT SOME THOUGHT If H2 is activated to produce H+, what must the other product be?
Hydrogen forms strong covalent bonds with many other elements, including oxygen; the O ¬ H bond enthalpy is 463 kJ>mol. The formation of the strong O ¬ H bond makes hydrogen an effective reducing agent for many metal oxides. When H2 is passed over heated CuO, for example, copper is produced: CuO(s) + H2(g) ¡ Cu(s) + H2O(g)
[22.6]
Hydrogen
921
922
CHAPTER 22
Chemistry of the Nonmetals
When H2 is ignited in air, a vigorous reaction occurs, forming H2O. Air containing as little as 4% H2 by volume is potentially explosive. Combustion of hydrogen-oxygen mixtures is used in liquid-fuel rocket engines such as those of the Space Shuttle. The hydrogen and oxygen are stored at low temperatures in liquid form.
Production of Hydrogen When a small quantity of H2 is needed in the laboratory, it is usually obtained by the reaction between an active metal such as zinc and a dilute strong acid such as HCl or H2SO4: Zn(s) + 2 H+(aq) ¡ Zn2+(aq) + H2(g)
[22.7]
Large quantities of H2 are produced by reacting methane with steam at 1100 °C. We can view this process as involving two reactions: CH4(g) + H2O(g) ¡ CO(g) + 3 H2(g)
[22.8]
CO(g) + H2O(g) ¡ CO2(g) + H2(g)
[22.9]
Carbon heated with water to about 1000 °C is another source of H2: C(s) + H2O(g) ¡ H2(g) + CO(g)
[22.10]
This mixture, known as water gas, is used as an industrial fuel.
A CLOSER LOOK THE HYDROGEN ECONOMY The reaction of hydrogen with oxygen is highly exothermic: 2 H2(g) + O2(g) ¡ 2 H2O(g) ¢H = -483.6 kJ
[22.11]
Because the only product of the reaction is water vapor, the prospect of using hydrogen as a fuel in fuel cells is attractive. •(Section 20.7) Alternatively, hydrogen could be combusted directly with oxygen from the atmosphere in an internal combustion engine. In either case, it would be necessary to generate elemental hydrogen on a large scale and arrange for its transport and storage.
Solar thermal
Elec tro ly s
Electrical nuclear
is
Natural gas
RELATED EXERCISES: 22.29, 22.30, 22.91
Hydroelectric, solar photovoltaics, wind
ewable energies Ren
Biomass
쑼 FIGURE 22.4 illustrates various sources and uses of H2 fuel. The generation of H2 through electrolysis of water is in principle the cleanest route, because this process—the reverse of Equation 22.11— produces only hydrogen and oxygen. •(Figure 1.7 and Section 20.9) However, the energy required to electrolyze water must come from somewhere. If we burn fossil fuels to generate this energy, we have not advanced very far toward a true hydrogen economy. If the energy for electrolysis came instead from a hydroelectric or nuclear power plant, solar cells, or wind generators, consumption of nonrenewable energy sources and undesired production of CO2 could be avoided.
Supply
Thermal nuclear
Coal
H2 Fuel cell engines
Demand
Processes, syntheses
t
B u i l di n g s
du
or
Com- Residential Other mercial
In
sp
str
y
n Tra Internal combustion engines
F u el c e lls
Turbines, internal combustion engines
Thermal
씱 FIGURE 22.4 The “hydrogen economy” would require hydrogen to be produced from various sources and would use hydrogen in energy-related applications.
SECTION 22.2
Hydrogen
923
Electrolysis of water consumes too much energy and is consequently too costly to be used commercially to produce H2. However, H2 is produced as a by-product in the electrolysis of brine (NaCl) solutions in the course of commercial Cl2 and NaOH manufacture: electrolysis
2 NaCl(aq) + 2 H2O(l) 999: H2(g) + Cl 2(g) + 2 NaOH(aq) [22.12] GIVE IT SOME THOUGHT What are the oxidation states of the H atoms in Equations 22.7–22.12?
Uses of Hydrogen Hydrogen is commercially important. About 5.0 * 1010 kg (50 million metric tons) is produced annually across the world. About half of the H2 produced is used to synthesize ammonia by the Haber process. • (Section 15.2) Much of the remaining hydrogen is used to convert high-molecular-weight hydrocarbons from petroleum into lowermolecular-weight hydrocarbons suitable for fuel (gasoline, diesel, and others) in a process known as cracking. Hydrogen is also used to manufacture methanol via the catalytic reaction of CO and H2 at high pressure and temperature: CO(g) + 2 H2(g) ¡ CH3OH(g)
[22.13]
Binary Hydrogen Compounds Hydrogen reacts with other elements to form three types of compounds: (1) ionic hydrides, (2) metallic hydrides, and (3) molecular hydrides. The ionic hydrides are formed by the alkali metals and by the heavier alkaline earths (Ca, Sr, and Ba). These active metals are much less electronegative than hydrogen. Consequently, hydrogen acquires electrons from them to form hydride ions (H-): Ca(s) + H2(g) ¡ CaH2(s)
[22.14]
The hydride ion is very basic and reacts readily with compounds having even weakly acidic protons to form H2: H-(aq) + H2O(l) ¡ H2(g) + OH-(aq)
[22.15]
Ionic hydrides can therefore be used as convenient (although expensive) sources of H2. Calcium hydride (CaH2) is used to inflate life rafts, weather balloons, and the like where a simple, compact means of generating H2 is desired (쑼 FIGURE 22.5).
GO FIGURE
This reaction is exothermic. Is the beaker on the right warmer or colder than the beaker on the left? CaH2
H2 gas H2O with pH indicator
Color change indicates presence of OH⫺
씱 FIGURE 22.5 The reaction of CaH2 with water.
924
CHAPTER 22
Chemistry of the Nonmetals
The reaction between H- and H2O (Equation 22.15) is an acid–base reaction and a redox reaction. The H- ion, therefore, is a good base and a good reducing agent. In fact, hydrides are able to reduce O2 to OH-: 2 NaH(s) + O2(g) ¡ 2 NaOH(s)
GO FIGURE
Which is the most thermodynamically stable hydride? Which is the least thermodynamically stable? 4A
5A
6A
7A
CH4(g) NH3(g) H2O(l) HF(g) ⫺50.8 ⫺16.7 ⫺237 ⫺271 SiH4(g) PH3(g) H2S(g) HCl(g) ⫹56.9 ⫹18.2 ⫺33.0 ⫺95.3 GeH4(g) AsH3(g) H2Se(g) HBr(g) ⫹117 ⫹111 ⫹71 ⫺53.2 SbH3(g) H2Te(g) HI(g) ⫹187 ⫹138 ⫹1.30 쑿 FIGURE 22.6 Standard free energies of formation of molecular hydrides. All values are kilojoules per mole of hydride.
[22.16]
For this reason, hydrides are normally stored in an environment that is free of both moisture and air. Metallic hydrides are formed when hydrogen reacts with transition metals. These compounds are so named because they retain their metallic properties. They are not molecular substances, just as metals are not. In many metallic hydrides, the ratio of metal atoms to hydrogen atoms is not fixed or in small whole numbers. The composition can vary within a range, depending on reaction conditions. TiH2 can be produced, for example, but preparations usually yield TiH1.8. These nonstoichiometric metallic hydrides are sometimes called interstitial hydrides. Because hydrogen atoms are small enough to fit between the sites occupied by the metal atoms, many metal hydrides behave like interstitial alloys. • (Section 12.3) The most interesting interstitial metallic hydride is that of palladium. Palladium can take up nearly 900 times its volume of hydrogen, making it very attractive for hydrogen storage in any possible future “hydrogen economy.” However, to be practical, any hydrogen-storage compound will have to contain 75% or more hydrogen by mass and be able to charge and discharge hydrogen quickly and safely near room temperature. GIVE IT SOME THOUGHT Palladium has a density of 12.023 g/cm3. Can a sample of Pd that has a volume of 1 cm3 increase its mass to over 900 g by adsorbing hydrogen?
The molecular hydrides, formed by nonmetals and metalloids, are either gases or liquids under standard conditions. The simple molecular hydrides are listed in 씱 FIGURE 22.6, together with their standard free energies of formation, ¢G f°. • (Section 19.5) In each family the thermal stability (measured as ¢G f°) decreases as we move down the family. (Recall that the more stable a compound is with respect to its elements under standard conditions, the more negative ¢G f° is.)
22.3 | GROUP 8A: THE NOBLE GASES 8A 2 He 10 Ne 18 Ar 36 Kr 54 Xe 86 Rn
The elements of group 8A are chemically unreactive. Indeed, most of our references to these elements have been in relation to their physical properties, as when we discussed intermolecular forces. • (Section 11.2) The relative inertness of these elements is due to the presence of a completed octet of valence-shell electrons (except He, which only has a filled 1s shell). The stability of such an arrangement is reflected in the high ionization energies of the group 8A elements. • (Section 7.4) The group 8A elements are all gases at room temperature. They are components of Earth’s atmosphere, except for radon, which exists only as a short-lived radioisotope. •(Section 21.9) Only argon is relatively abundant. • (Table 18.1) Neon, argon, krypton, and xenon are recovered from liquid air by distillation. All four of these noble gases are used in lighting, display, and laser applications in which the atoms are excited electrically and electrons that are in a higher energy state emit light as they fall to the ground state. • (Section 6.2) Argon is used as a blanketing atmosphere in electric lightbulbs. The gas conducts heat away from the filament but does not react with it. Argon is also used as a protective atmosphere to prevent oxidation in welding and certain hightemperature metallurgical processes. Helium is in many ways the most important noble gas. Liquid helium is used as a coolant to conduct experiments at very low temperatures. Helium boils at 4.2 K and 1 atm, the lowest boiling point of any substance. It is found in relatively high concentrations in many natural-gas wells and can be obtained from them.
SECTION 22.3
Group 8A: The Noble Gases
925
Noble-Gas Compounds Because the noble gases are exceedingly stable, they react only under rigorous conditions. We expect the heavier ones to be most likely to form compounds because their ionization energies are lower. • (Figure 7.9) A lower ionization energy suggests the possibility of sharing an electron with another atom, leading to a chemical bond. In addition, because the group 8A elements (except helium) already contain eight electrons in their valence shell, formation of covalent bonds will require an expanded valence shell. Valence-shell expansion occurs most readily with larger atoms. • (Section 8.7) The first noble-gas compound was reported in 1962. This discovery caused a sensation because it undercut the belief that the noble-gas elements were inert. The initial study involved xenon in combination with fluorine, the element we would expect to be most reactive in pulling electron density from another atom. Since that time chemists have prepared several xenon compounds of fluorine and oxygen (쑼 TABLE 22.1). The fluorides XeF2, XeF4, and XeF6 are made by direct reaction of the elements. By varying the ratio of reactants and altering reaction conditions, one of the three compounds can be obtained. The oxygen-containing compounds are formed when the fluorides react with water as, for example, XeF6(s) + 3 H2O(l) ¡ XeO3(aq) + 6 HF(aq) SAMPLE EXERCISE 22.3
[22.17]
Predicting a Molecular Structure
Use the VSEPR model to predict the structure of XeF4. SOLUTION Analyze We must predict the geometrical structure given only the molecular formula. Plan We must first write the Lewis structure for the molecule. We then count the number of electron pairs (domains) around the Xe atom and use that number and the number of bonds to predict the geometry. Solve There are 36 valence-shell electrons (8 from xenon and 7 from each fluorine). If we make four single Xe¬F bonds, each fluorine has its octet satisfied. Xe then has 12 electrons in its valence shell, so we expect an octahedral arrangement of six electron pairs. Two of these are nonbonded pairs. Because nonbonded pairs require more volume than bonded pairs •(Section 9.2), it is reasonable to expect these nonbonded pairs to be opposite each other. The expected structure is square planar, as shown in 씰 FIGURE 22.7. Comment The experimentally determined structure agrees with this prediction. PRACTICE EXERCISE Describe the electron-domain geometry and molecular geometry of XeF2. Answer: trigonal bipyramidal, linear
The other noble-gas elements form compounds much less readily than xenon. For many years, only one binary krypton compound, KrF2, was known with certainty, and it decomposes to its elements at -10 °C. Other compounds of krypton have been isolated at very low temperatures (40 K). TABLE 22.1 • Properties of Xenon Compounds Compound
Oxidation State of Xe
Melting Point (°C)
¢H f° (kJ/mol)a
+2 +4 +6 +6 +6 +6 +8
129 117 49 -41 to -28
-109(g) -218(g) -298(g) +146(l) +402(s) +145(s) —
XeF2 XeF4 XeF6 XeOF4 XeO3 XeO2F2 XeO4 a
At 25 °C, for the compound in the state indicated.
b
A solid; decomposes at 40 °C. A solid; decomposes at -40 °C.
c
–b 31 –c
F
F Xe
F
F
쑿 FIGURE 22.7 Xenon tetrafluoride.
926
CHAPTER 22
Chemistry of the Nonmetals
22.4 | GROUP 7A: THE HALOGENS 7A 9 F 17 Cl 35 Br 53 I 85 At
The elements of group 7A, the halogens, have the outer-electron configuration ns2np5, where n ranges from 2 through 6. The halogens have large negative electron affinities • (Section 7.5), and they most often achieve a noble-gas configuration by gaining an electron, which results in a -1 oxidation state. Fluorine, being the most electronegative element, exists in compounds only in the -1 state. The other halogens exhibit positive oxidation states up to +7 in combination with more electronegative atoms such as O. In the positive oxidation states, the halogens tend to be good oxidizing agents, readily accepting electrons. Chlorine, bromine, and iodine are found as the halides in seawater and in salt deposits. Fluorine occurs in the minerals fluorspar (CaF2), cryolite (Na3AlF6), and fluorapatite [Ca5(PO4)3F].* Only fluorspar is an important commercial source of fluorine. All isotopes of astatine are radioactive. The longest-lived isotope is astatine-210, which has a half-life of 8.1 hr and decays mainly by electron capture. Because astatine is so unstable, very little is known about its chemistry.
GO FIGURE
Do Br2 and I2 appear to be more or less soluble in CCl4 than in H2O? Cl2(aq) added
NaF(aq)
NaBr(aq)
NaI(aq)
CCl4 layer No reaction
I⫺ Br⫺ oxidized oxidized to Br2 to I2
쑿 FIGURE 22.8 Reaction of Cl2 with aqueous solutions of NaF, NaBr, and NaI. The top liquid layer is water; the bottom liquid layer is carbon tetrachloride.
Properties and Production of the Halogens Most properties of the halogens vary in a regular fashion as we go from fluorine to iodine (쑼 TABLE 22.2). Under ordinary conditions the halogens exist as diatomic molecules. The molecules are held together in the solid and liquid states by dispersion forces. • (Section 11.2) Because I2 is the largest and most polarizable halogen molecule, the intermolecular forces between I2 molecules are the strongest. Thus, I2 has the highest melting point and boiling point. At room temperature and 1 atm, I2 is a purple solid, Br2 is a red-brown liquid, and Cl2 and F2 are gases. • (Figure 7.27) Chlorine readily liquefies upon compression at room temperature and is normally stored and handled in liquid form under pressure in steel containers. The comparatively low bond enthalpy of F2 (155 kJ> mol) accounts in part for the extreme reactivity of elemental fluorine. Because of its high reactivity, F2 is difficult to work with. Certain metals, such as copper and nickel, can be used to contain F2 because their surfaces form a protective coating of metal fluoride. Chlorine and the heavier halogens are also reactive, although less so than fluorine. Because of their high electronegativities, the halogens tend to gain electrons from other substances and thereby serve as oxidizing agents. The oxidizing ability of the halogens, indicated by their standard reduction potentials, decreases going down the group. As a result, a given halogen is able to oxidize the halide anions below it. For example, Cl2 oxidizes Br - and I - but not F - , as seen in 씱 FIGURE 22.8.
TABLE 22.2 • Some Properties of the Halogens Property Atomic radius (Å) Ionic radius, X- (Å) First ionization energy (kJ>mol) Electron affinity (kJ>mol) Electronegativity X ¬ X single-bond enthalpy (kJ>mol) Reduction potential (V): 1 2 X2(aq) + e ¡ X (aq)
F
Cl
Br
I
0.71 1.33 1681 -328 4.0 155
0.99 1.81 1251 -349 3.0 242
1.14 1.96 1140 -325 2.8 193
1.33 2.20 1008 -295 2.5 151
2.87
1.36
1.07
0.54
*Minerals are solid substances that occur in nature. They are usually known by their common names rather than by their chemical names. What we know as rock is merely an aggregate of different minerals.
SECTION 22.4
SAMPLE EXERCISE 22.4
Group 7A: The Halogens
927
Predicting Chemical Reactions among the Halogens
Write the balanced equation for the reaction, if any, between (a) I -(aq) and Br2(l), (b) Cl-(aq) and I2(s). SOLUTION Analyze We are asked to determine whether a reaction occurs when a particular halide and halogen are combined. Plan A given halogen is able to oxidize anions of the halogens below it in the periodic table. Thus, in each pair the halogen having the smaller atomic number ends up as the halide ion. If the halogen with the smaller atomic number is already the halide, there is no reaction. Thus, the key to determining whether a reaction occurs is locating the elements in the periodic table. Solve (a) Br2 can oxidize (remove electrons from) the anions of the halogens below it in the periodic table. Thus, it oxidizes I - : 2 I -(aq) + Br2(aq) ¡ I2(s) + 2 Br -(aq) (b) Cl- is the anion of a halogen above iodine in the periodic table. Thus, I2 cannot oxidize Cl- ; there is no reaction. PRACTICE EXERCISE Write the balanced chemical equation for the reaction between Br -(aq) and Cl2(aq). Answer: 2 Br -(aq) + Cl 2(aq) ¡ Br2(aq) + 2 Cl-(aq)
Notice in Table 22.2 that the standard reduction potential of F2 is exceptionally high. Fluorine gas readily oxidizes water: F2(aq) + H2O(l) ¡ 2 HF(aq) +
1 2
O2(g)
E° = 1.80 V
[22.18]
Fluorine cannot be prepared by electrolytic oxidation of aqueous solutions of fluoride salts because water is oxidized more readily than F -. • (Section 20.9) In practice, the element is formed by electrolytic oxidation of a solution of KF in anhydrous HF. Chlorine is produced mainly by electrolysis of either molten or aqueous sodium chloride. Both bromine and iodine are obtained commercially from brines containing the halide ions; the reaction used is oxidation with Cl2.
Uses of the Halogens Fluorine is used to prepare fluorocarbons—very stable carbon–fluorine compounds used as refrigerants, lubricants, and plastics. Teflon® (씰 FIGURE 22.9) is a polymeric fluorocarbon noted for its high thermal stability and lack of chemical reactivity. Chlorine is by far the most commercially important halogen. About 1 * 1010 kg (10 million tons) of Cl2 is produced annually in the United States. In addition, hydrogen chloride production is about 4.0 * 109 kg (4.4 million tons) annually. About half of this chlorine finds its way eventually into the manufacture of chlorine-containing organic compounds, such as the vinyl chloride (C2H3Cl) used in making polyvinyl chloride (PVC) plastics. • (Section 12.8) Much of the remainder is used as a bleaching agent in the paper and textile industries. When Cl2 dissolves in cold dilute base, it converts into Cl- and hypochlorite, ClO- : Cl 2(aq) + 2 OH-(aq) Δ Cl-(aq) + ClO-(aq) + H2O(l)
[22.19]
GIVE IT SOME THOUGHT What is the oxidation state of Cl in each Cl species in Equation 22.19?
Sodium hypochlorite (NaClO) is the active ingredient in many liquid bleaches. Chlorine is also used in water treatment to oxidize and thereby destroy bacteria. • (Section 18.4)
GO FIGURE
What is the repeating unit in this polymer?
쑿 FIGURE 22.9 Structure of Teflon®, a fluorocarbon polymer.
928
CHAPTER 22
Chemistry of the Nonmetals
A common use of iodine is as KI in table salt. Iodized salt provides the small amount of iodine necessary in our diets; it is essential for the formation of thyroxin, a hormone secreted by the thyroid gland. Lack of iodine in the diet results in an enlarged thyroid gland, a condition called goiter.
The Hydrogen Halides All the halogens form stable diatomic molecules with hydrogen. Aqueous solutions of HCl, HBr, and HI are strong acids. The hydrogen halides can be formed by direct reaction of the elements. The most important means of preparing HF and HCl, however, is by reacting a salt of the halide with a strong nonvolatile acid, as in the reaction ¢
CaF2(s) + H2SO4(l) ¡ 2 HF(g) + CaSO4(s)
[22.20]
Neither HBr nor HI can be prepared in this way, however, because H2SO4 oxidizes Br - and I - (쑼 FIGURE 22.10). This difference in reactivity reflects the greater ease of oxidation of Br - and I - relative to F - and Cl-. These undesirable oxidations are avoided by using a nonvolatile acid, such as H3PO4, that is a weaker oxidizing agent than H2SO4. SAMPLE EXERCISE 22.5
Writing a Balanced Chemical Equation
Write a balanced equation for the formation of hydrogen bromide gas from the reaction of solid sodium bromide with phosphoric acid. SOLUTION Analyze We are asked to write a balanced equation for the reaction between NaBr and H3PO4 to form HBr and another product. Plan As in Equation 22.20, a metathesis reaction takes place. • (Section 4.2) Let’s assume that only one H in H3PO4 reacts. (The actual number depends on the reaction conditions.) The H2PO4 - and Na + will form NaH2PO4 as one product. Solve The balanced equation is NaBr(s) + H3PO4(l) ¡ NaH2PO4(s) + HBr(g) PRACTICE EXERCISE Write the balanced equation for the preparation of HI from NaI and H3PO4. Answer: NaI(s) + H3PO4(l) ¡ NaH2PO4(s) + HI(g)
GO FIGURE
Are these reactions acid–base reactions or oxidation-reduction reactions? H2SO4
씰 FIGURE 22.10 Reaction of H2SO4 with NaI and NaBr.
NaI
NaBr
I2 formed
Br2 formed
SECTION 22.4
Group 7A: The Halogens
929
TABLE 22.3 • The Stable Oxyacids of the Halogens Oxidation State of Halogen +1 +3 +5 +7
Formula of Acid Cl
Br
I
Acid Name
HClO HClO2 HClO3 HClO4
HBrO — HBrO3 HBrO4
HIO — HIO3 HIO4
Hypohalous acid Halous acid Halic acid Perhalic acid
The hydrogen halides form hydrohalic acid solutions when dissolved in water. These solutions have the characteristic properties of acids, such as reactions with active metals to produce hydrogen gas. • (Section 4.4) Hydrofluoric acid also reacts readily with silica (SiO2) and with silicates to form hexafluorosilicic acid (H2SiF6): SiO2(s) + 6 HF(aq) ¡ H2SiF6(aq) + 2 H2O(l)
[22.21]
Interhalogen Compounds Because the halogens exist as diatomic molecules, diatomic molecules made up of two different halogen atoms exist. These compounds are the simplest examples of interhalogens, compounds, such as ClF and IF5, formed between two halogen elements. With one exception, the higher interhalogen compounds have a central Cl, Br, or I atom surrounded by 3, 5, or 7 fluorine atoms. The large size of the iodine atom allows the formation of IF3, IF5, and IF7, in which the oxidation state of I is +3, +5, and +7, respectively. With the smaller bromine and chorine atoms, only compounds with 3 or 5 fluorines form. The only higher interhalogen compounds that do not have outer F atoms are ICl3 and ICl5; the large size of the I atom can accommodate 5 Cl atoms, whereas Br is not large enough to allow even BrCl3 to form. All of the interhalogen compounds are powerful oxidizing agents.
Oxyacids and Oxyanions
10 Al(s) ⫹ 6 NH4ClO4(s) 4 Al2O3(s) ⫹ 2 AlCl3(s)
⫹ 12 H2O(g) ⫹ 3 N2(g)
쑿 TABLE 22.3 summarizes the formulas of
the known oxyacids of the halogens and the way they are named.* • (Section 2.8) The acid strengths of the oxyacids increase with increasing oxidation state of the central halogen atom. • (Section 16.10) All the oxyacids are strong oxidizing agents. The oxyanions, formed on removal of H+ from the oxyacids, are generally more stable than the oxyacids. Hypochlorite salts are used as bleaches and disinfectants because of the powerful oxidizing capabilities of the ClOion. Sodium hypochlorite is used as a bleaching agent. Chlorate salts are similarly very reactive. For example, potassium chlorate is used to make matches and fireworks. GIVE IT SOME THOUGHT Which do you expect to be the stronger oxidizing agent, NaBrO3 or NaClO3?
Perchloric acid and its salts are the most stable oxyacids and oxyanions. Dilute solutions of perchloric acid are quite safe, and many perchlorate salts are stable except when heated with organic materials. When heated, however, perchlorates can become vigorous, even violent, oxidizers. Considerable caution should be exercised, therefore, when handling these substances, and it is crucial to avoid contact between perchlorates and readily oxidized material. The use of ammonium perchlorate (NH4ClO4) as the oxidizer in the solid booster rockets for the Space Shuttle demonstrates the oxidizing power of perchlorates. The solid propellant contains a mixture of NH4ClO4 and powdered aluminum, the reducing agent. Each shuttle launch requires about 6 * 105 kg (700 tons) of NH4ClO4 (씰 FIGURE 22.11). *Fluorine forms one oxyacid, HOF. Because the electronegativity of fluorine is greater than that of oxygen, we must consider fluorine to be in a -1 oxidation state and oxygen to be in the 0 oxidation state in this compound.
The large volume of gases produced provides thrust for the booster rockets 쑿 FIGURE 22.11 Launch of the Space Shuttle Columbia from the Kennedy Space Center.
930
CHAPTER 22
Chemistry of the Nonmetals
CHEMISTRY AND LIFE HOW MUCH PERCHLORATE IS TOO MUCH? Since the 1950s both NASA and the Pentagon have used ammonium perchlorate, NH4ClO4, as a rocket fuel. The result is that traces of perchlorate ion are found in groundwater in many regions of the United States, with levels ranging from about 4 to 100 ppb. Perchlorate is known to suppress thyroid hormone levels in humans. However, it is disputed whether the amounts found in drinking water are sufficiently high to cause health problems. The Environmental Protection Agency currently states that a dose of
0.007 mg per kilogram of body mass per day is not expected to cause adverse health effects in humans. For a 70-kg (154-lb.) person drinking 2 L of water per day, that amounts to a concentration of 25 ppb. California has proposed a standard of 6 ppb. Removal of perchlorate ion from water supplies is not easy. Although perchlorate is an oxidizing agent, the ClO4 - ion is quite stable in aqueous solution. One promising avenue is reduction by microorganisms. While research continues on the best means of decreasing perchlorate levels in drinking water, federal agencies continue to explore what level constitutes a safe upper limit. RELATED EXERCISE: 22.93
22.5 | OXYGEN
쑿 FIGURE 22.12 Oxygen is the name of a play by the chemists Carl Djerassi (Stanford University) and Roald Hoffmann (Cornell University). Its subject is the controversy over who discovered oxygen.
By the middle of the seventeenth century, scientists recognized that air contained a component associated with burning and breathing. That component was not isolated until 1774, however, when English scientist Joseph Priestley discovered oxygen. Lavoisier subsequently named the element oxygen, meaning “acid former.” There is some historical debate about who “really” discovered oxygen, and the debate is the subject of a play written by the chemist-authors Carl Djerassi and Roald Hoffmann (씱 FIGURE 22.12). Oxygen is found in combination with other elements in a great variety of compounds—water (H2O), silica (SiO2), alumina (Al2O3), and the iron oxides (Fe2O3, Fe3O4) are obvious examples. Indeed, oxygen is the most abundant element by mass both in Earth’s crust and in the human body. • (Section 1.2) It is the oxidizing agent for the metabolism of our foods and is crucial to human life.
Properties of Oxygen Oxygen has two allotropes, O2 and O3. When we speak of molecular oxygen or simply oxygen, it is usually understood that we are speaking of dioxygen (O2), the normal form of the element; O3 is ozone. At room temperature, dioxygen is a colorless, odorless gas. It condenses to a liquid at -183 °C and freezes at -218 °C. It is only slightly soluble in water (0.04 g> L, or 0.001 M at 25 °C), but its presence in water is essential to marine life. The electron configuration of the oxygen atom is 3He42s22p4. Thus, oxygen can complete its octet of valence electrons either by picking up two electrons to form the oxide ion (O2-) or by sharing two electrons. In its covalent compounds, it tends to form either two single bonds, as in H2O, or a double bond, as in formaldehyde (H2C “ O). The O2 molecule contains a double bond. The bond in O2 is very strong (bond enthalpy 495 kJ> mol). Oxygen also forms strong bonds with many other elements. Consequently, many oxygencontaining compounds are thermodynamically more stable than O2. In the absence of a catalyst, however, most reactions of O2 have high activation energies and thus require high temperatures to proceed at a suitable rate. Once a sufficiently exothermic reaction begins, it may accelerate rapidly, producing a reaction of explosive violence.
Production of Oxygen Nearly all commercial oxygen is obtained from air. The normal boiling point of O2 is -183 °C, whereas that of N2, the other principal component of air, is -196 °C. Thus, when air is liquefied and then allowed to warm, the N2 boils off, leaving liquid O2 contaminated mainly by small amounts of N2 and Ar. In the laboratory, O2 can be obtained by heating either aqueous hydrogen peroxide or solid potassium chlorate (KClO3): 2 KClO3(s) ¡ 2 KCl(s) + 3 O2(g) Manganese dioxide (MnO2) catalyzes both reactions.
[22.22]
SECTION 22.5
Much of the O2 in the atmosphere is replenished through photosynthesis, in which green plants use the energy of sunlight to generate O2 (along with glucose, C6H12O6) from atmospheric CO2:
Oxygen
2 C2H2(g) ⫹ 5 O2(g) 4 CO2(g) ⫹ 2 H2O(g); Δ H° ⫽ ⫺2510 kJ
6 CO2(g) + 6 H2O(l) ¡ C6H12O6(aq) + 6 O2(g)
Uses of Oxygen In industrial use, oxygen ranks behind only sulfuric acid (H2SO4) and nitrogen (N2). About 3 * 1010 kg (30 million tons) of O2 is used annually in the United States. It is shipped and stored either as a liquid or in steel containers as a compressed gas. About 70% of the O2 output, however, is generated where it is needed. Oxygen is by far the most widely used oxidizing agent in industry. Over half of the O2 produced is used in the steel industry, mainly to remove impurities from steel. It is also used to bleach pulp and paper. (Oxidation of colored compounds often gives colorless products.) Oxygen is used together with acetylene (C2H2) in oxyacetylene welding (씰 FIGURE 22.13). The reaction between C2H2 and O2 is highly exothermic, producing temperatures in excess of 3000 °C. 쑿 FIGURE 22.13 Welding with an oxyacetylene torch.
Ozone Ozone is a pale blue, poisonous gas with a sharp, irritating odor. Most people can detect about 0.01 ppm in air. Exposure to 0.1 to 1 ppm produces headaches, burning eyes, and irritation to the respiratory passages. The O3 molecule possesses a p bond that is delocalized over the three oxygen atoms. • (Section 8.6) The molecule dissociates readily, forming reactive oxygen atoms: O3(g) ¡ O2(g) + O(g)
¢H° = 105 kJ
[22.23]
GIVE IT SOME THOUGHT What wavelength of light is needed to break an O¬O bond in ozone?
Ozone is a stronger oxidizing agent than dioxygen. Ozone forms oxides with many elements under conditions where O2 will not react; indeed, it oxidizes all the common metals except gold and platinum. Ozone can be prepared by passing electricity through dry O2 in a flow-through apparatus. The electrical discharge causes the O2 bond to break, resulting in reactions like those described in Section 18.1. During thunderstorms, ozone is generated (and can be smelled, if you are too close) from lightning strikes: electricity
3 O2(g) 999: 2 O3(g)
¢H ⴰ = 285 kJ
[22.24]
Ozone cannot be stored for long, except at low temperature, because it readily decomposes to O2. The decomposition is catalyzed by Ag, Pt, and Pd and by many transition-metal oxides. SAMPLE EXERCISE 22.6
Calculating an Equilibrium Constant
Using ¢G f° for ozone from Appendix C, calculate the equilibrium constant for Equation 22.24 at 298.0 K, assuming no electrical input. SOLUTION Analyze We are asked to calculate the equilibrium constant for the formation of O3 from O2, given the temperature and ¢G f°. Plan The relationship between the standard free-energy change, ¢G f°, for a reaction and the equilibrium constant for the reaction is given in Equation 19.20. Solve From Appendix C we have
¢G f° (O3) = 163.4 kJ>mol
Thus, for Equation 22.24,
¢G° = (2 mol O3)(163.4 kJ>mol O3) = 326.8 kJ
From Equation 19.20 we have
¢G° = -RT ln K
Thus,
ln K =
-326.8 * 103 J - ¢G° = = -131.9 RT (8.314 J>K-mol)(298.0 K) K = e - 131.9 = 5.207 * 10 - 58
931
932
CHAPTER 22
Chemistry of the Nonmetals
Comment In spite of the unfavorable equilibrium constant, ozone can be prepared from O2 as described in the preceding text. The unfavorable free energy of formation is overcome by energy from the
electrical discharge, and O3 is removed before the reverse reaction can occur, so a nonequilibrium mixture results.
PRACTICE EXERCISE Using the data in Appendix C, calculate ¢G° and K for Equation 22.23 at 298.0 K. Answer: ¢G° = 66.7 kJ, K = 2.03 * 10 - 12
Ozone is sometimes used to treat drinking water. Like Cl2, ozone kills bacteria and oxidizes organic compounds. The largest use of ozone, however, is in the preparation of pharmaceuticals, synthetic lubricants, and other commercially useful organic compounds, where O3 is used to sever carbon–carbon double bonds. Ozone is an important component of the upper atmosphere, where it screens out ultraviolet radiation and so protects us from the effects of these high-energy rays. For this reason, depletion of stratospheric ozone is a major scientific concern. • (Section 18.2) In the lower atmosphere, ozone is considered an air pollutant and is a major constituent of smog. • (Section 18.2) Because of its oxidizing power, ozone damages living systems and structural materials, especially rubber.
Oxides The electronegativity of oxygen is second only to that of fluorine. As a result, oxygen has negative oxidation states in all compounds except OF2 and O2F2. The -2 oxidation state is by far the most common. Compounds in this oxidation state are called oxides. Nonmetals form covalent oxides, most of which are simple molecules with low melting and boiling points. SiO2 and B2O3, however, have extended structures. Most nonmetal oxides combine with water to give oxyacids. Sulfur dioxide (SO2), for example, dissolves in water to give sulfurous acid (H2SO3): SO2(g) + H2O(l) ¡ H2SO3(aq)
[22.25]
This reaction and that of SO3 with H2O to form H2SO4 are largely responsible for acid rain. • (Section 18.2) The analogous reaction of CO2 with H2O to form carbonic acid (H2CO3) causes the acidity of carbonated water. Oxides that form acids when they react with water are called either acidic anhydrides (anhydride means “without water”) or acidic oxides. A few nonmetal oxides, especially ones with the nonmetal in a low oxidation state—such as N2O, NO, and CO—do not react with water and are not acidic anhydrides. GIVE IT SOME THOUGHT What acid is produced by the reaction of I2O5 with water?
Most metal oxides are ionic compounds. The ionic oxides that dissolve in water form hydroxides and, consequently, are called either basic anhydrides or basic oxides. Barium oxide, for example, reacts with water to form barium hydroxide (씰 FIGURE 22.14). These kinds of reactions are due to the high basicity of the O2- ion and its virtually complete hydrolysis in water: O2-(aq) + H2O(l) ¡ 2 OH-(aq) TABLE 22.4 • Acid–Base Character of Chromium Oxides Oxide
Oxidation State of Cr
CrO Cr2O3 CrO3
+2 +3 +6
Nature of Oxide Basic Amphoteric Acidic
[22.26]
Even those ionic oxides that are insoluble in water tend to dissolve in strong acids. Iron(III) oxide, for example, dissolves in acids: Fe 2O3(s) + 6 H+(aq) ¡ 2 Fe 3+(aq) + 3 H2O(l)
[22.27]
This reaction is used to remove rust (Fe 2O3 # nH2O) from iron or steel before a protective coat of zinc or tin is applied. Oxides that can exhibit both acidic and basic character are said to be amphoteric. • (Section 17.5) If a metal forms more than one oxide, the basic character of the oxide decreases as the oxidation state of the metal increases (씱 TABLE 22.4).
SECTION 22.5
Oxygen
933
GO FIGURE
Is this reaction a redox reaction?
H2O with indicator
4 KO2(s) ⫹ 2 H2O(l, from breath) 4 K⫹(aq) ⫹ 4 OH⫺(aq) ⫹ 3 O2(g) 2 OH⫺(aq) ⫹ CO2(g, from breath) H2O(l) ⫹ CO32⫺(aq) Pink color indicates basic solution
BaO(s)
ⴙ
H2O(l)
Ba(OH)2(aq)
쑿 FIGURE 22.14 Reaction of a basic oxide with water.
Peroxides and Superoxides Compounds containing O ¬ O bonds and oxygen in the -1 oxidation state are peroxides. Oxygen has an oxidation state of - 12 in O2 - , which is called the superoxide ion. The most active (easily oxidized) metals (K, Rb, and Cs) react with O2 to give superoxides (KO2, RbO2, and CsO2). Their active neighbors in the periodic table (Na, Ca, Sr, and Ba) react with O2, producing peroxides (Na2O2, CaO2, SrO2, and BaO2). Less active metals and nonmetals produce normal oxides. • (Section 7.6) When superoxides dissolve in water, O2 is produced: +
-
4 KO2(s) + 2 H2O(l) ¡ 4 K (aq) + 4 OH (aq) + 3 O2(g)
[22.28]
Because of this reaction, potassium superoxide is used as an oxygen source in masks worn by rescue workers (씰 FIGURE 22.15). For proper breathing in toxic environments, oxygen must be generated in the mask and exhaled carbon dioxide in the mask must be eliminated. Moisture in the breath causes the KO2 to decompose to O2 and KOH, and the KOH removes CO2 from the exhaled breath: -
2 OH (aq) + CO2(g) ¡ H2O(l) + CO3 (aq) 2-
[22.29]
쑿 FIGURE 22.15 A self-contained breathing apparatus.
GO FIGURE
Does H2O2 have a dipole moment?
Hydrogen peroxide (씰 FIGURE 22.16) is the most familiar and commercially important peroxide. Pure hydrogen peroxide is a clear, syrupy liquid that melts at -0.4 °C. Concentrated hydrogen peroxide is dangerously reactive because the decomposition to water and oxygen is very exothermic: 2 H2O2(l) ¡ 2 H2O(l) + O2(g)
¢H° = -196.1 kJ
[22.30]
This is another example of a disproportionation reaction, in which an element is simultaneously oxidized and reduced. The oxidation number of oxygen changes from -1 to -2 and 0. Hydrogen peroxide is marketed as a chemical reagent in aqueous solutions of up to about 30% by mass. A solution containing about 3% H2O2 by mass is sold in drugstores and used as a mild antiseptic. Somewhat more concentrated solutions are used to bleach fabrics. The peroxide ion is a by-product of metabolism that results from the reduction of O2. The body disposes of this reactive ion with enzymes such as peroxidase and catalase.
94⬚
쑿 FIGURE 22.16 Molecular structure of hydrogen peroxide.
934
CHAPTER 22
Chemistry of the Nonmetals
|
22.6 THE OTHER GROUP 6A ELEMENTS: S, Se, Te, AND Po 6A 8 O 16 S 34 Se 52 Te 84 Po
The other group 6A elements are sulfur, selenium, tellurium, and polonium. In this section we will survey the properties of the group as a whole and then examine the chemistry of sulfur, selenium, and tellurium. We will not say much about polonium, which has no stable isotopes and is found only in minute quantities in radium-containing minerals.
General Characteristics of the Group 6A Elements The group 6A elements possess the general outer-electron configuration ns2np4 with n ranging from 2 to 6. Thus, these elements attain a noble-gas electron configuration by adding two electrons, which results in a -2 oxidation state. Except for oxygen, the group 6A elements are also commonly found in positive oxidation states up to +6, and they can have expanded valence shells. Thus, we have such compounds as SF6, SeF6, and TeF6 with the central atom in the +6 oxidation state. 쑼 TABLE 22.5 summarizes some properties of the group 6A elements.
Occurrence and Production of S, Se, and Te Sulfur, selenium, and tellurium can all be mined from the earth. Large underground deposits are the principal source of elemental sulfur (씱 FIGURE 22.17). Sulfur also occurs widely as sulfide (S2-) and sulfate (SO42- ) minerals. Its presence as a minor component of coal and petroleum poses a major problem. Combustion of these “unclean” fuels leads to serious pollution by sulfur oxides. •(Section 18.2) Much effort has been directed at removing this sulfur, and these efforts have increased the availability of sulfur. The sale of this sulfur helps to partially offset the costs of the desulfurizing processes and equipment. Selenium and tellurium occur in rare minerals, such as Cu2Se, PbSe, Cu2Te, and PbTe, and as minor constituents in sulfide ores of copper, iron, nickel, and lead. 쑿 FIGURE 22.17 Massive amounts of sulfur are extracted every year from the earth.
Properties and Uses of Sulfur, Selenium, and Tellurium Elemental sulfur is yellow, tasteless, and nearly odorless. It is insoluble in water and exists in several allotropic forms. The thermodynamically stable form at room temperature is rhombic sulfur, which consists of puckered S8 rings. • (Figure 7.26) When heated above its melting point (113 °C), molten sulfur first contains S8 molecules and is fluid because the rings slip over one another. Further heating causes the rings to break; the fragments then join to form long molecules that can become entangled. In this polymeric form, sulfur becomes highly viscous. Further heating breaks the chains, and the viscosity again decreases. Most of the approximately 1 * 1010 kg (10 million tons) of sulfur produced in the United States each year is used to manufacture sulfuric acid. Sulfur is also used to vulcanize rubber, a process that toughens rubber by introducing cross-linking between polymer chains. • (Section 12.8) TABLE 22.5 • Some Properties of the Group 6A Elements Property Atomic radius (Å) X2- ionic radius (Å) First ionization energy (kJ>mol) Electron affinity (kJ>mol) Electronegativity X ¬ X single-bond enthalpy (kJ>mol) Reduction potential to H2X in acidic solution (V) *Based on O ¬ O bond energy in H2O2.
O 0.73 1.40 1314 -141 3.5 146* 1.23
S
Se
Te
1.04 1.84 1000 -200 2.5 266
1.17 1.98 941 -195 2.4 172
1.43 2.21 869 -190 2.1 126
0.14
-0.40
-0.72
SECTION 22.6
The Other Group 6A Elements: S, Se, Te, and Po
Selenium and tellurium do not form eight-membered rings in their elemental forms. • (Section 7.8) The most stable allotropes of these elements are crystalline substances containing helical chains of atoms (씰 FIGURE 22.18). Each atom is close to atoms in adjacent chains, and it appears that some sharing of electron pairs between these atoms occurs. The electrical conductivity of elemental selenium is low in the dark but increases greatly upon exposure to light. This property is exploited in photoelectric cells and light meters. Photocopiers also depend on the photoconductivity of selenium. Photocopy machines contain a belt or drum coated with a film of selenium. This drum is electrostatically charged and then exposed to light reflected from the image being photocopied. The charge drains from the regions where the selenium film has been made conductive by exposure to light. A black powder (the toner) sticks only to the areas that remain charged. The photocopy is made when the toner is transferred to a sheet of plain paper.
Se
Se
Se
Se
Se
Se Se
Se
Se
Se
Se
Se Se
935
Se
쑿 FIGURE 22.18 Portion of helical chains making up the structure of crystalline selenium.
Sulfides When an element is less electronegative than sulfur, sulfides that contain S2- form. Many metallic elements are found in the form of sulfide ores, such as PbS (galena) and HgS (cinnabar). A series of related ores containing the disulfide ion, S2 2- (analogous to the peroxide ion), are known as pyrites. Iron pyrite, FeS2, occurs as golden yellow cubic crystals (씰 FIGURE 22.19). Because it has been occasionally mistaken for gold by miners, iron pyrite is often called fool’s gold. One of the most important sulfides is hydrogen sulfide (H2S). This substance is not normally produced by direct union of the elements because it decomposes at elevated temperatures. It is normally prepared by action of dilute acid on iron(II) sulfide: FeS(s) + 2 H+(aq) ¡ H2S(aq) + Fe 2+(aq)
[22.31]
One of hydrogen sulfide’s most readily recognized properties is its odor, which is most frequently encountered as the offensive odor of rotten eggs. Hydrogen sulfide is toxic but our noses can detect H2S in extremely low, nontoxic concentrations. A sulfurcontaining organic molecule, such as dimethyl sulfide, (CH3)2S, which is similarly odoriferous and can be detected by smell at a level of one part per trillion, is added to natural gas as a safety factor to give it a detectable odor.
Oxides, Oxyacids, and Oxyanions of Sulfur Sulfur dioxide, formed when sulfur burns in air, has a choking odor and is poisonous. The gas is particularly toxic to lower organisms, such as fungi, so it is used to sterilize dried fruit and wine. At 1 atm and room temperature, SO2 dissolves in water to produce a 1.6 M solution. The SO2 solution is acidic, and we describe it as sulfurous acid (H2SO3). Salts of SO3 2- (sulfites) and HSO3 - (hydrogen sulfites or bisulfites) are well known. Small quantities of Na2SO3 or NaHSO3 are used as food additives to prevent bacterial spoilage. Because some people are extremely allergic to sulfites, all food products with sulfites must now carry a warning label disclosing their presence. Although combustion of sulfur in air produces mainly SO2, small amounts of SO3 are also formed. The reaction produces chiefly SO2 because the activationenergy barrier for oxidation to SO3 is very high unless the reaction is catalyzed. Interestingly, the SO3 by-product is used industrially to make H2SO4, which is the ultimate product of the reaction between SO3 and water. In the manufacture of sulfuric acid, SO2 is obtained by burning sulfur and then oxidized to SO3, using a catalyst such as V2O5 or platinum. The SO3 is dissolved in H2SO4 because it does not dissolve quickly in water, and then the H2S2O7 formed in this reaction, called pyrosulfuric acid, is added to water to form H2SO4: SO3(g) + H2SO4(l) ¡ H2S2O7(l)
[22.32]
H2S2O7(l) + H2O(l) ¡ 2 H2SO4(l)
[22.33]
쑿 FIGURE 22.19 Iron pyrite (FeS2, on the right) with gold for comparison.
936
CHAPTER 22
Chemistry of the Nonmetals
GO FIGURE
In this reaction, what has happened to the H and O atoms in the sucrose? H2SO4
씰 FIGURE 22.20 Sulfuric acid dehydrates table sugar to produce elemental carbon.
Table sugar (sucrose), C12H22O11
Pure carbon, C
GIVE IT SOME THOUGHT What is the net reaction of Equations 22.32 and 22.33?
Commercial sulfuric acid is 98% H2SO4. It is a dense, colorless, oily liquid that boils at 340 °C. It is a strong acid, a good dehydrating agent (쑿 FIGURE 22.20), and a moderately good oxidizing agent. Year after year, the production of sulfuric acid is the largest of any chemical produced in the United States. About 4 * 1010 kg (40 million tons) is produced annually in this country. Sulfuric acid is employed in some way in almost all manufacturing. Consequently, its consumption is considered a measure of industrial activity. Sulfuric acid is a strong acid, but only the first hydrogen is completely ionized in aqueous solution: H2SO4(aq) ¡ H+(aq) + HSO4 -(aq) -
+
HSO4 (aq) Δ H (aq) + SO4 (aq) Ka = 1.1 * 10 2-
[22.34] -2
[22.35] (SO42-
GO FIGURE
What are the oxidation states of the sulfur atoms in the S2O32 ⴚ ion? 2⫺
2⫺
Consequently, sulfuric acid forms both sulfates salts) and bisulfates (or hydrogen sulfates, HSO4 salts). Bisulfate salts are common components of the “dry acids” used for adjusting the pH of swimming pools and hot tubs; they are also components of many toilet bowl cleaners. The term thio indicates substitution of sulfur for oxygen, and the thiosulfate ion (S2O3 2-) is formed by boiling an alkaline solution of SO3 2- with elemental sulfur: 8 SO3 2-(aq) + S8(s) ¡ 8 S2O3 2-(aq)
쑿 FIGURE 22.21 Structures of the sulfate (left) and thiosulfate (right) ions.
[22.36]
The structures of the sulfate and thiosulfate ions are compared in 씱 FIGURE 22.21. Thiosulfate salts are used industrially in the paper-making and textile industries and are the “fixer” solutions in the development of photographic film.
SECTION 22.7
Nitrogen
937
22.7 | NITROGEN In 1772 the Scottish botanist Daniel Rutherford found that a mouse enclosed in a sealed jar quickly consumed the life-sustaining component of air (oxygen) and died. When the “fixed air” (CO2) in the container was removed, a “noxious air” remained that would not sustain combustion or life. We now know that gas as nitrogen. Nitrogen constitutes 78% by volume of Earth’s atmosphere, where it occurs as N2 molecules. Although nitrogen is a key element in living organisms, compounds of nitrogen are not abundant in Earth’s crust. The major natural deposits of nitrogen compounds are those of KNO3 (saltpeter) in India and NaNO3 (Chile saltpeter) in Chile and other desert regions of South America.
Properties of Nitrogen Nitrogen is a colorless, odorless, tasteless gas composed of N2 molecules. Its melting point is -210 °C, and its normal boiling point is -196 °C. The N2 molecule is very unreactive because of the strong triple bond between nitrogen atoms (the N ‚ N bond enthalpy is 941 kJ> mol, nearly twice that for the bond in O2; • Table 8.4). When substances burn in air, they normally react with O2 but not with N2. When magnesium burns in air, however, it reacts with N2 to form magnesium nitride (Mg3N2): 3 Mg(s) + N2(g) ¡ Mg 3N2(s)
[22.37]
A similar reaction occurs with lithium, forming Li3N. The nitride ion is a strong Brønsted–Lowry base and reacts with water to form ammonia: Mg 3N2(s) + 6 H2O(l) ¡ 2 NH3(aq) + 3 Mg(OH)2(s)
[22.38]
The electron configuration of the nitrogen atom is 3He42s 2p . The element exhibits all formal oxidation states from +5 to -3 (씰 TABLE 22.6). The +5, 0, and -3 oxidation states are the most common and generally the most stable of these. Because nitrogen is more electronegative than all other elements except fluorine, oxygen, and chlorine, it exhibits positive oxidation states only in combination with these three elements. 2
TABLE 22.6 • Oxidation States of Nitrogen Oxidation State +5 +4 +3 +2 +1
3
Production and Uses of Nitrogen Elemental nitrogen is obtained in commercial quantities by fractional distillation of liquid air. About 4 * 1010 kg (40 million tons) of N2 is produced annually in the United States. Because of its low reactivity, large quantities of N2 are used as an inert gaseous blanket to exclude O2 in food processing, manufacture of chemicals, metal fabrication, and production of electronic devices. Liquid N2 is employed as a coolant to freeze foods rapidly. The largest use of N2 is in the manufacture of nitrogen-containing fertilizers, which provide a source of fixed nitrogen. We have previously discussed nitrogen fixation in the “Chemistry and Life” box in Section 14.7 and in the “Chemistry Put to Work” box in Section 15.2. Our starting point in fixing nitrogen is the manufacture of ammonia via the Haber process. • (Section 15.2) The ammonia can then be converted into a variety of useful, simple nitrogen-containing species (씰 FIGURE 22.22).
Hydrogen Compounds of Nitrogen Ammonia is one of the most important compounds of nitrogen. It is a colorless, toxic gas that has a characteristic irritating odor. As noted in previous discussions, the NH3 molecule is basic (Kb = 1.8 * 10 - 5). • (Section 16.7)
0 -1 -2 -3
Examples N2O5, HNO3, NO3 NO2, N2O4 HNO2, NO2 - , NF3 NO N2O, H2N2O2, N2O2 2- , HNF2 N2 NH2OH, NH2F N2H4 NH3, NH4 + , NH2 -
938
CHAPTER 22
Chemistry of the Nonmetals
GO FIGURE
In which of these species is the oxidation number of nitrogen ⴙ3? N2H4 (hydrazine)
N2 (dinitrogen)
NH3 (ammonia)
NO (nitric oxide)
NH4ⴙ (ammonium salts)
NO2 (nitrogen dioxide)
NO2ⴚ (nitrite salts)
HNO3 (nitric acid)
NO3ⴚ (nitrate salts)
쑿 FIGURE 22.22 Sequence of conversion of N2 into common nitrogen compounds.
GO FIGURE
Is the N¬N bond length in these molecules shorter or longer than the N¬N bond length in N2?
In the laboratory, NH3 can be prepared by the action of NaOH on an ammonium salt. The NH4 + ion, which is the conjugate acid of NH3, transfers a proton to OH- . The resultant NH3 is volatile and is driven from the solution by mild heating: NH4Cl(aq) + NaOH(aq) ¡ NH3(g) + H2O(l) + NaCl(aq)
[22.39]
Commercial production of NH3 is achieved by the Haber process: N2(g) + 3 H2(g) ¡ 2 NH3(g)
[22.40]
About 1 * 1010 kg (10 million tons) of ammonia is produced annually in the United States. About 75% is used for fertilizer. Hydrazine (N2H4) is another important hydride of nitrogen. The hydrazine molecule contains an N ¬ N single bond (씱 FIGURE 22.23). Hydrazine is quite poisonous. It can be prepared by the reaction of ammonia with hypochlorite ion (OCl-) in aqueous solution: 2 NH3(aq) + OCl-(aq) ¡ N2H4(aq) + Cl-(aq) + H2O(l) 쑿 FIGURE 22.23 Hydrazine (top, N2H4) and methylhydrazine (bottom, CH3NHNH2).
SAMPLE EXERCISE 22.7
The reaction involves several intermediates, including chloramine (NH2Cl). The poisonous NH2Cl bubbles out of solution when household ammonia and chlorine bleach (which contains OCl-) are mixed. This reaction is one reason for the frequently cited warning not to mix bleach and household ammonia. Pure hydrazine is a strong and versatile reducing agent. The major use of hydrazine and compounds related to it, such as methylhydrazine (Figure 22.23), is as rocket fuel.
Writing a Balanced Equation
Hydroxylamine (NH2OH) reduces copper(II) to the free metal in acid solutions. Write a balanced equation for the reaction, assuming that N2 is the oxidation product. SOLUTION Analyze We are asked to write a balanced oxidation-reduction equation in which NH2OH is converted to N2 and Cu2+ is converted to Cu. Plan Because this is a redox reaction, the equation can be balanced by the method of half-reactions discussed in Section 20.2. Thus, we begin with two half-reactions, one involving the NH2OH and N2 and the other involving Cu2+ and Cu. Solve The unbalanced and incomplete half-reactions are
[22.41]
Cu2+(aq) ¡ Cu(s) NH2OH(aq) ¡ N2(g)
SECTION 22.7
Balancing these equations as described in Section 20.2 gives Adding these half-reactions gives the balanced equation:
Nitrogen
939
Cu2+(aq) + 2 e - ¡ Cu(s) 2 NH2OH(aq) ¡ N2(g) + 2 H2O(l) + 2 H+(aq) + 2 e Cu2+(aq) + 2 NH2OH(aq) ¡ Cu(s) + N2(g) + 2 H2O(l) + 2 H+(aq)
PRACTICE EXERCISE (a) In power plants, hydrazine is used to prevent corrosion of the metal parts of steam boilers by the O2 dissolved in the water. The hydrazine reacts with O2 in water to give N2 and H2O. Write a balanced equation for this reaction. (b) Methylhydrazine, N2H3CH3(l), is used
with the oxidizer dinitrogen tetroxide, N2O4(l), to power the steering rockets of the Space Shuttle orbiter. The reaction of these two substances produces N2, CO2, and H2O. Write a balanced equation for this reaction.
Answers: (a) N2H4(aq) + O2(aq) ¡ N2(g) + 2 H2O(l);
(b) 5 N2O4(l) + 4 N2H3CH3(l) ¡ 9 N2(g) + 4 CO2(g) + 12 H2O(g)
Oxides and Oxyacids of Nitrogen Nitrogen forms three common oxides: N2O (nitrous oxide), NO (nitric oxide), and NO2 (nitrogen dioxide). It also forms two unstable oxides that we will not discuss, N2O3 (dinitrogen trioxide) and N2O5 (dinitrogen pentoxide). Nitrous oxide (N2O) is also known as laughing gas because a person becomes giddy after inhaling a small amount. This colorless gas was the first substance used as a general anesthetic. It is used as the compressed gas propellant in several aerosols and foams, such as in whipped cream. It can be prepared in the laboratory by carefully heating ammonium nitrate to about 200 °C: ¢
NH4NO3(s) ¡ N2O(g) + 2 H2O(g)
[22.42]
Nitric oxide (NO) is also a colorless gas but, unlike N2O, it is slightly toxic. It can be prepared in the laboratory by reduction of dilute nitric acid, using copper or iron as a reducing agent (쑼 FIGURE 22.24): 3 Cu(s) + 2 NO3 -(aq) + 8 H+(aq) ¡ 3 Cu2+(aq) + 2 NO(g) + 4 H2O(l) [22.43] Nitric oxide is also produced by direct reaction of N2 and O2 at high temperatures. This reaction is a significant source of nitrogen oxide air pollutants. • (Section 18.2) The direct combination of N2 and O2 is not used for commercial production of NO, however, because the yield is low, the equilibrium constant Kp at 2400 K being only 0.05. • (Section 15.7, “Chemistry Put to Work: Controlling Nitric Oxide Emissions”) The commercial route to NO (and hence to other oxygen-containing compounds of nitrogen) is via the catalytic oxidation of NH3: Pt catalyst
4 NH3(g) + 5 O2(g) ——¡ 4 NO(g) + 6 H2O(g) 850 °C
[22.44]
This reaction is the first step in the Ostwald process, by which NH3 is converted commercially into nitric acid (HNO3).
HNO3(aq)
Cu metal
NO(g)
Cu2⫹(aq) solution 씱 FIGURE 22.24 Copper metal reacts with nitric acid to produce blue Cu2 + (aq) and NO(g).
940
Chemistry of the Nonmetals
CHAPTER 22
NO2(g) NO(g)
O2 in air
쑿 FIGURE 22.25 Formation of NO2(g) as NO(g) combines with O2(g) in the air.
When exposed to air, nitric oxide reacts readily with O2 (쑿 FIGURE 22.25): 2 NO(g) + O2(g) ¡ 2 NO2(g)
[22.45]
When dissolved in water, NO2 forms nitric acid: 3 NO2(g) + H2O(l) ¡ 2 H+(aq) + 2 NO3 -(aq) + NO(g)
[22.46]
Nitrogen is both oxidized and reduced in this reaction, which means it disproportionates. The NO can be converted back into NO2 by exposure to air (Equation 22.45) and thereafter dissolved in water to prepare more HNO3. NO is an important neurotransmitter in the human body. It causes the muscles that line blood vessels to relax, thus allowing an increased passage of blood (see the “Chemistry and Life” box on page 941). Nitrogen dioxide (NO2) is a yellow-brown gas (Figure 22.25). Like NO, it is a major constituent of smog. • (Section 18.2) It is poisonous and has a choking odor. As discussed in Section 15.1, NO2 and N2O4 exist in equilibrium: 2 NO2(g) Δ N2O4(g) GO FIGURE
Which is the shortest NO bond in these two molecules? O H
H
O
O
N
N
O
O
쑿 FIGURE 22.26 Structures of nitric acid (top) and nitrous acid (bottom).
¢H° = -58 kJ
[22.47]
The two common oxyacids of nitrogen are nitric acid (HNO3) and nitrous acid (HNO2) (씱 FIGURE 22.26). Nitric acid is a strong acid. It is also a powerful oxidizing agent, as indicated by the standard reduction potential in the reaction NO3 -(aq) + 4 H+(aq) + 3 e - ¡ NO(g) + 2 H2O(l)
E° = +0.96 V
[22.48]
Concentrated nitric acid attacks and oxidizes most metals except Au, Pt, Rh, and Ir. About 7 * 109 kg (8 million tons) of nitric acid is produced annually in the United States. Its largest use is in the manufacture of NH4NO3 for fertilizers. It is also used in the production of plastics, drugs, and explosives. Among the explosives made from nitric acid are nitroglycerin, trinitrotoluene (TNT), and nitrocellulose. The following reaction occurs when nitroglycerin explodes: 4 C3H5N3O9(l) ¡ 6 N2(g) + 12 CO2(g) + 10 H2O(g) + O2(g)
[22.49]
All the products of this reaction contain very strong bonds and are gases. As a result, the reaction is very exothermic, and the volume of the products is far larger than the volume occupied by the reactant. Thus, the expansion resulting from the heat generated by the reaction produces the explosion. • (Section 8.8: “Chemistry Put to Work: Explosives and Alfred Nobel”) Nitrous acid is considerably less stable than HNO3 and tends to disproportionate into NO and HNO3. It is normally made by action of a strong acid, such as H2SO4, on a cold solution of a nitrite salt, such as NaNO2. Nitrous acid is a weak acid (Ka = 4.5 * 10-4). GIVE IT SOME THOUGHT What are the oxidation numbers of the nitrogen atoms in a. nitric acid; b. nitrous acid?
SECTION 22.8
The Other Group 5A Elements: P, As, Sb, and Bi
941
CHEMISTRY AND LIFE NITROGLYCERIN AND HEART DISEASE
blood vessels. In 1998 the Nobel Prize in Physiology or Medicine was awarded to Robert F. Furchgott, Louis J. Ignarro, and Ferid Murad for their discoveries of the detailed pathways by which NO acts in the cardiovascular system. It was a sensation that this simple, common air pollutant could exert important functions in mammals, including humans. As useful as nitroglycerin is to this day in treating angina pectoris, it has a limitation in that prolonged administration results in development of tolerance, or desensitization, of the vascular muscle to further vasorelaxation by nitroglycerin. The bioactivation of nitroglycerin is the subject of active research in the hope that a means of circumventing desensitization can be found.
During the 1870s an interesting observation was made in Alfred Nobel’s dynamite factories. Workers who suffered from heart disease that caused chest pains when they exerted themselves found relief from the pains during the workweek. It quickly became apparent that nitroglycerin, present in the air of the factory, acted to enlarge blood vessels. Thus, this powerfully explosive chemical became a standard treatment for angina pectoris, the chest pains accompanying heart failure. It took more than 100 years to discover that nitroglycerin was converted in the vascular smooth muscle into NO, which is the chemical agent actually causing dilation of the
|
22.8 THE OTHER GROUP 5A ELEMENTS: P, As, Sb, AND Bi Of the other group 5A elements—phosphorus, arsenic, antimony, and bismuth— phosphorus has a central role in several aspects of biochemistry and environmental chemistry.
5A 7 N
General Characteristics of the Group 5A Elements
15 P
The group 5A elements have the outer-shell electron configuration ns2np3, with n ranging from 2 to 6. A noble-gas configuration is achieved by adding three electrons to form the -3 oxidation state. Ionic compounds containing X3- ions are not common, however. More commonly, the group 5A element acquires an octet of electrons via covalent bonding and oxidation numbers ranging from -3 to +5. Because of its lower electronegativity, phosphorus is found more frequently in positive oxidation states than is nitrogen. Furthermore, compounds in which phosphorus has the +5 oxidation state are not as strongly oxidizing as the corresponding compounds of nitrogen. Compounds in which phosphorus has a -3 oxidation state are much stronger reducing agents than are the corresponding nitrogen compounds. Some properties of the group 5A elements are listed in 쑼 TABLE 22.7. The general pattern is similar to what we saw with other groups: Size and metallic character increase as atomic number increases in the group. The variation in properties among group 5A elements is more striking than that seen in groups 6A and 7A. Nitrogen at the one extreme exists as a gaseous diatomic molecule, clearly nonmetallic. At the other extreme, bismuth is a reddish white, metallic-looking substance that has most of the characteristics of a metal. The values listed for X ¬ X single-bond enthalpies are not reliable because it is difficult to obtain such data from thermochemical experiments. However, there is no doubt
TABLE 22.7 • Properties of the Group 5A Elements Property Atomic radius (Å) First ionization energy (kJ>mol) Electron affinity (kJ>mol) Electronegativity X ¬ X single-bond enthalpy (kJ>mol)* X ‚ X triple-bond enthalpy (kJ>mol) *Approximate values only.
N
P
As
Sb
Bi
0.75 1402 7 0 3.0 163
1.10 1012 -72 2.1 200
1.21 947 -78 2.0 150
1.41 834 -103 1.9 120
1.55 703 -91 1.9 —
380
295
192
941
490
33 As 51 Sb 83 Bi
942
CHAPTER 22
Chemistry of the Nonmetals
about the general trend: a low value for the N ¬ N single bond, an increase at phosphorus, and then a gradual decline to arsenic and antimony. From observations of the elements in the gas phase, it is possible to estimate the X ‚ X triple-bond enthalpies. Here we see a trend that is different from that for the X ¬ X single bond. Nitrogen forms a much stronger triple bond than the other elements, and there is a steady decline in the triplebond enthalpy down through the group. These data help us to appreciate why nitrogen alone of the group 5A elements exists as a diatomic molecule in its stable state at 25 °C. All the other elements exist in structural forms with single bonds between the atoms.
Occurrence, Isolation, and Properties of Phosphorus Phosphorus occurs mainly in the form of phosphate minerals. The principal source of phosphorus is phosphate rock, which contains phosphate principally as Ca3(PO4)2. The element is produced commercially by reduction of calcium phosphate with carbon in the presence of SiO2: 1500 °C
2 Ca3(PO4)2(s) + 6 SiO2(s) + 10 C(s) 999: P4(g) + 6 CaSiO3(l) + 10 CO(g) [22.50]
White phosphorus
The phosphorus produced in this fashion is the allotrope known as white phosphorus. This form distills from the reaction mixture as the reaction proceeds. Phosphorus exists in several allotropic forms. White phosphorus consists of P4 tetrahedra (씱 FIGURE 22.27). The bond angles in this molecule, 60°, are unusually small, so there is much strain in the bonding, which is consistent with the high reactivity of white phosphorus. This allotrope bursts spontaneously into flames if exposed to air. When heated in the absence of air to about 400 °C, white phosphorus is converted to a more stable allotrope known as red phosphorus, which does not ignite on contact with air. Red phosphorus is Red phosphorus also considerably less poisonous than the white form. We will denote elemental phosphorus as simply P(s).
쑿 FIGURE 22.27 White and red phosphorus. Despite the fact that both contain nothing but phosphorus atoms, these two forms of phosphorus differ greatly in reactivity. The white allotrope, which reacts violently with oxygen, must be stored under water so that it is not exposed to air. The much less reactive red form does not need to be stored this way.
Phosphorus Halides Phosphorus forms a wide range of compounds with the halogens, the most important of which are the trihalides and pentahalides. Phosphorus trichloride (PCl3) is commercially the most significant of these compounds and is used to prepare a wide variety of products, including soaps, detergents, plastics, and insecticides. Phosphorus chlorides, bromides, and iodides can be made by direct oxidation of elemental phosphorus with the elemental halogen. PCl3, for example, which is a liquid at room temperature, is made by passing a stream of dry chlorine gas over white or red phosphorus: 2 P(s) + 3 Cl 2(g) ¡ 2 PCl 3(l)
[22.51]
If excess chlorine gas is present, an equilibrium is established between PCl3 and PCl5. PCl 3(l) + Cl 2(g) Δ PCl 5(s)
[22.52]
The phosphorus halides hydrolyze on contact with water. The reactions occur readily, and most of the phosphorus halides fume in air because of reaction with water vapor. In the presence of excess water the products are the corresponding phosphorus oxyacid and hydrogen halide: PBr3(l) + 3 H2O(l) ¡ H3PO3(aq) + 3 HBr(aq)
[22.53]
PCl 5(l) + 4 H2O(l) ¡ H3PO4(aq) + 5 HCl(aq)
[22.54]
Oxy Compounds of Phosphorus Probably the most significant phosphorus compounds are those in which the element is combined with oxygen. Phosphorus(III) oxide (P4O6) is obtained by allowing white phosphorus to oxidize in a limited supply of oxygen. When oxidation takes place in the
SECTION 22.8
The Other Group 5A Elements: P, As, Sb, and Bi
presence of excess oxygen, phosphorus(V) oxide (P4O10) forms. This compound is also readily formed by oxidation of P4O6. These two oxides represent the two most common oxidation states for phosphorus, +3 and +5. The structural relationship between P4O6 and P4O10 is shown in 씰 FIGURE 22.28. Notice the resemblance these molecules have to the P4 molecule (Figure 22.27); all three substances have a P4 core.
GO FIGURE
How do the electron domains about P in P4O6 differ from those about P in P4O10? O
SAMPLE EXERCISE 22.8
943
P
Calculating a Standard Enthalpy Change
The reactive chemicals on the tip of a “strike anywhere” match are usually P4S3 and an oxidizing agent such as KClO3. When the match is struck on a rough surface, the heat generated by the friction ignites the P4S3, and the oxidizing agent brings about rapid combustion. The products of the combustion of P4S3 are P4O10 and SO2. Calculate the standard enthalpy change for the combustion of P4S3 in air, given the following standard enthalpies of formation: P4S3 1 -154.4 kJ>mol2, P4O10 (-2940.1 kJ>mol), SO2 (-296.9 kJ>mol). SOLUTION Analyze We are given the reactants (P4S3 and O2 from air) and the products (P4O10 and SO2) for a reaction, together with their standard enthalpies of formation, and asked to calculate the standard enthalpy change for the reaction. Plan We first need a balanced chemical equation for the reaction. The enthalpy change for the reaction is then equal to the enthalpies of formation of products minus those of reactants (Equation 5.31). We also need to recall that the standard enthalpy of formation of any element in its standard state is zero. Thus, ¢H f°(O2) = 0. Solve The balanced chemical equation for the combustion is P4S3(s) + 8 O2(g) ¡ P4O10(s) + 3 SO2(g) Thus, we can write ¢H° = ¢H f°(P4O10) + 3 ¢H f°(SO2) - ¢H f°(P4S3) - 8 ¢H f°(O2) = -2940.1 kJ + 3(-296.9) kJ - (-154.4 kJ) - 8(0)
쑿 FIGURE 22.28 Structures of P4O6 (top) and P4O10 (bottom).
= -3676.4 kJ Comment The reaction is strongly exothermic, making it evident why P4S3 is used on match tips. PRACTICE EXERCISE Write the balanced equation for the reaction of P4O10 with water, and calculate ¢H° for this reaction using data from Appendix C. Answer: P4O10(s) + 6 H2O(l) ¡ 4 H3PO4(aq); -498.1 kJ
Phosphorus(V) oxide is the anhydride of phosphoric acid (H3PO4), a weak triprotic acid. In fact, P4O10 has a very high affinity for water and is consequently used as a drying agent. Phosphorus(III) oxide is the anhydride of phosphorous acid (H3PO3), a weak diprotic acid (씰 FIGURE 22.29).* One characteristic of phosphoric and phosphorous acids is their tendency to undergo condensation reactions when heated. • (Section 12.8) For example, two H3PO4 molecules are joined by the elimination of one H2O molecule to form H4P2O7: O HO
P OH
O OH ⫹ HO
P
O
O OH
HO
OH
P OH
O
P
OH ⫹ H2O
OH
This H not acidic because P–H bond is nonpolar
These atoms are eliminated as H2O
[22.55] Phosphoric acid and its salts find their most important uses in detergents and fertilizers. The phosphates in detergents are often in the form of sodium tripolyphosphate (Na5P3O10). *Note that the element phosphorus (FOS # for # us) has a -us suffix, whereas the first word in the name phosphorous (fos # FOR # us) acid has an -ous suffix.
쑿 FIGURE 22.29 Structures of H3PO4 (top) and H3PO3 (bottom).
944
CHAPTER 22
Chemistry of the Nonmetals
The phosphate ions “soften” water by binding their oxygen groups to the metal ions that contribute to the hardness of water. This keeps the metal ions from interfering with the action of the detergent. The phosphates also keep the pH above 7 and thus prevent the detergent molecules from becoming protonated. Most mined phosphate rock is converted to fertilizers. The Ca3(PO4)2 in phosphate rock is insoluble (Ksp = 2.0 * 10-29). It is converted to a soluble form for use in fertilizers by treatment with sulfuric or phosphoric acid. The reaction with phosphoric acid yields Ca(H2PO4)2: Ca3(PO4)2(s) + 4 H3PO4(aq) ¡ 3 Ca2+(aq) + 6 H2PO4 -(aq)
[22.56]
Although the solubility of Ca(H2PO4)2 allows it to be assimilated by plants, it also allows it to be washed from the soil and into bodies of water, thereby contributing to water pollution. • (Section 18.4) Phosphorus compounds are important in biological systems. The element occurs in phosphate groups in RNA and DNA, the molecules responsible for the control of protein biosynthesis and transmission of genetic information. It also occurs in adenosine triphosphate (ATP), which stores energy in biological cells and has the structure NH2 N HC N O ⫺
O
O O
P ⫺
C
C
N
N CH
O O
P
O
P
⫺
O
C
CH2
⫺
O
O
O
C H
H C
H C
C H
OH
OH
Adenosine
The P ¬ O ¬ P bond of the end phosphate group is broken by hydrolysis with water, forming adenosine diphosphate (ADP): O ⫺
O
P O⫺
O O
P
O O
P
O⫺
Adenosine ⫹ H2O
O
O⫺
ATP O
O HO
P
O
O⫺
P O⫺
O O
Adenosine ⫹
⫺
O
P
OH [22.57]
O⫺
ADP This reaction releases 33 kJ of energy under standard conditions, but in the living cell, the Gibbs free energy change for the reaction is closer to -57 kJ> mol. The concentration of ATP inside a living cell is in the range of 1–10 mM, which means a typical human metabolizes her or his body mass of ATP in one day! ATP is continually made from ADP and continually converted back to ADP, releasing energy that can be harnessed by other cellular reactions.
SECTION 22.9
Carbon
945
CHEMISTRY AND LIFE ARSENIC IN DRINKING WATER “Arsenic,” meaning its oxides, has been known as a poison for centuries. The current Environmental Protection Agency (EPA) standard for arsenic in public water supplies is 10 ppb (equivalent to 10 mg>L). Most regions of the United States tend to have low to moderate (2–10 ppb) groundwater arsenic levels (쑼 FIGURE 22.30). The western region tends to have higher levels, coming mainly from natural geological sources in the area. Estimates, for example, indicate that 35% of water-supply wells in Arizona have arsenic concentrations above 10 ppb. The problem of arsenic in drinking water in the United States is dwarfed by the problem in other parts of the world—especially in Bangladesh, where the problem is tragic. Historically, surface water
Alaska Hawaii
EXPLANATION
Puerto Rico
At least 25% of samples exceed 50 ppm As At least 25% of samples exceed 10 ppm As At least 25% of samples exceed 5 ppm As At least 25% of samples exceed 3 ppm As At least 25% of samples exceed 1 ppm As Insufficient data
쑿 FIGURE 22.30 Geographic distribution of arsenic in groundwater.
sources in that country have been contaminated with microorganisms, causing significant health problems, including one of the highest infant mortality rates in the world. During the 1970s, international agencies, headed by the United Nations Children’s Fund (UNICEF), began investing millions of dollars of aid money in Bangladesh for wells to provide “clean” drinking water. Unfortunately, no one tested the well water for the presence of arsenic; the problem was not discovered until the 1980s. The result has been the biggest outbreak of mass poisoning in history. Up to half of the country’s estimated 10 million wells have arsenic concentrations above 50 ppb. In water the most common forms of arsenic are the arsenate ion and its protonated hydrogen anions (AsO4 3- , HAsO4 2- , and H2AsO4 - ) and the arsenite ion and its protonated forms (AsO3 3- , HAsO3 2- , H2AsO3 - , and H3AsO3). These species are collectively referred to by the oxidation number of the arsenic as arsenic(V) and arsenic(III), respectively. Arsenic(V) is more prevalent in oxygenrich (aerobic) surface waters, whereas arsenic(III) is more likely to occur in oxygen-poor (anaerobic) groundwaters. In the pH range from 4 to 10, the arsenic(V) is present primarily as HAsO4 2- and H2AsO4 - , and the arsenic(III) is present primarily as the neutral acid H3AsO3. One of the challenges in determining the health effects of arsenic in drinking waters is the different chemistry of arsenic(V) and arsenic(III), as well as the different concentrations required for physiological responses in different individuals. In Bangladesh, skin lesions were the first sign of the arsenic problem. Statistical studies correlating arsenic levels with the occurrence of disease indicate a lung and bladder cancer risk arising from even low levels of arsenic. The current technologies for removing arsenic perform most effectively when treating arsenic in the form of arsenic(V), so water treatment strategies require preoxidation of the drinking water. Once in the form of arsenic(V), there are a number of possible removal strategies. For example, Fe2(SO4)3 could be added to precipitate FeAsO4, which is then removed by filtration. RELATED EXERCISE: 22.104.
22.9 | CARBON Carbon constitutes only 0.027% of Earth’s crust. Although some carbon occurs in elemental form as graphite and diamond, most is found in combined form. Over half occurs in carbonate compounds, and carbon is also found in coal, petroleum, and natural gas. The importance of the element stems in large part from its occurrence in all living organisms: Life as we know it is based on carbon compounds.
Elemental Forms of Carbon We have seen that carbon exists in several allotropic crystalline forms: graphite, diamond, fullerenes, carbon nanotubes, and graphene. Fullerenes, nanotubes, and graphene are discussed in Chapter 12; here we focus on graphite and diamond. Graphite is a soft, black, slippery solid that has a metallic luster and conducts electricity. It consists of parallel sheets of sp2-hybridized carbon atoms held together by
946
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Chemistry of the Nonmetals
dispersion forces. • (Section 12.7) Diamond is a clear, hard solid in which the carbon atoms form an sp3-hybridized covalent network. • (Section 12.7) Diamond is denser than graphite (d = 2.25 g>cm3 for graphite; d = 3.51 g>cm3 for diamond). At approximately 100,000 atm at 3000 °C, graphite converts to diamond. In fact, almost any carbon-containing substance, if put under sufficiently high pressure, forms diamonds; scientists at General Electric in the 1950s used peanut butter to make diamonds. About 3 * 104 kg of industrial-grade diamonds are synthesized each year, mainly for use in cutting, grinding, and polishing tools. Graphite has a well-defined crystalline structure, but it also exists in two common amorphous forms: carbon black and charcoal. Carbon black is formed when hydrocarbons are heated in a very limited supply of oxygen, such as in this methane reaction: CH4(g) + O2(g) ¡ C(s) + 2 H2O(g)
[22.58]
Carbon black is used as a pigment in black inks; large amounts are also used in making automobile tires. Charcoal is formed when wood is heated strongly in the absence of air. Charcoal has an open structure, giving it an enormous surface area per unit mass. “Activated charcoal,” a pulverized form of charcoal whose surface is cleaned by heating with steam, is widely used to adsorb molecules. It is used in filters to remove offensive odors from air and colored or bad-tasting impurities from water.
Oxides of Carbon Carbon forms two principal oxides: carbon monoxide (CO) and carbon dioxide (CO2). Carbon monoxide is formed when carbon or hydrocarbons are burned in a limited supply of oxygen: 2 C(s) + O2(g) ¡ 2 CO(g)
[22.59]
CO is a colorless, odorless, tasteless gas that is toxic because it binds to hemoglobin in the blood and thus interfere with oxygen transport. Low-level poisoning results in headaches and drowsiness; high-level poisoning can cause death. Automobile engines produce carbon monoxide, which is a major air pollutant. Carbon monoxide is unusual in that it has a nonbonding pair of electrons on carbon: C O . It is isoelectronic with N2, so you might expect CO to be equally unreactive. Moreover, both substances have high bond energies (1072 kJ> mol for C ‚ O and 941 kJ> mol for N ‚ N). Because of the lower nuclear charge on carbon (compared with either N or O), however, the carbon nonbonding pair is not held as strongly as that on N or O. Consequently, CO is better able to function as a Lewis base than is N2; for example, CO can coordinate its nonbonding pair to the iron of hemoglobin, displacing O2, but N2 cannot. In addition, CO forms a variety of covalent compounds, known as metal carbonyls, with transition metals. Ni(CO)4, for example, is a volatile, toxic compound formed by warming metallic nickel in the presence of CO. The formation of metal carbonyls is the first step in the transition-metal catalysis of a variety of reactions of CO. Carbon monoxide has several commercial uses. Because it burns readily, forming CO2, it is employed as a fuel: 2 CO(g) + O2(g) ¡ 2 CO2(g)
¢H° = - 566 kJ
[22.60]
Carbon monoxide is an important reducing agent, widely used in metallurgical operations to reduce metal oxides, such as the iron oxides: Fe 3O4(s) + 4 CO(g) ¡ 3 Fe(s) + 4 CO2(g)
[22.61]
Carbon dioxide is produced when carbon-containing substances are burned in excess oxygen, such as in this reaction involving ethanol: C2H5OH(l) + 3 O2(g) ¡ 2 CO2(g) + 3 H2O(g)
[22.62]
SECTION 22.9
Carbon
947
CHEMISTRY PUT TO WORK Carbon Fibers and Composites The properties of graphite are anisotropic; that is, they differ in different directions through the solid. Along the carbon planes, graphite possesses great strength because of the number and strength of the carbon–carbon bonds in this direction. The bonds between planes are relatively weak, however, making graphite weak in that direction. Fibers of graphite can be prepared in which the carbon planes are aligned to varying extents parallel to the fiber axis. These fibers are lightweight (density of about 2 g>cm3) and chemically quite unreactive. The oriented fibers are made by first slowly pyrolyzing (decomposing by action of heat) organic fibers at about 150 °C to 300 °C. These fibers are then heated to about 2500 °C to graphitize them (convert amorphous carbon to graphite). Stretching the fiber during pyrolysis helps orient the graphite planes parallel to the fiber axis. More amorphous carbon fibers are formed by pyrolysis of organic fibers at lower temperatures (1200 °C to 400 °C). These amorphous materials, commonly called carbon fibers, are the type most often used in commercial materials. Composite materials that take advantage of the strength, stability, and low density of carbon fibers are widely used. Composites are combinations of two or more materials. These materials are present as separate phases and are combined to form structures that take advantage of certain desirable properties of each component. In carbon composites the graphite fibers are often woven into a fabric that is embedded in a matrix that binds them into a solid structure. The
fibers transmit loads evenly throughout the matrix. The finished composite is thus stronger than any one of its components. Carbon composite materials are used widely in a number of applications, including high-performance graphite sports equipment such as tennis racquets, golf clubs, and bicycle wheels (쑼 FIGURE 22.31). Heat-resistant composites are required for many aerospace applications, where carbon composites now find wide use.
쑿 FIGURE 22.31 Carbon composites in commercial products.
It is also produced when many carbonates are heated: ¢
CaCO3(s) ¡ CaO(s) + CO2(g)
[22.63]
Large quantities of CO2 are obtained as a by-product of the fermentation of sugar during the production of ethanol: yeast
C6H12O6(aq) ¡ 2 C2H5OH(aq) + 2 CO2(g) Glucose
[22.64]
Ethanol
In the laboratory, CO2 can be produced by the action of acids on carbonates (씰 FIGURE 22.32): CO3 2-(aq) + 2 H+(aq) ¡ CO2(g) + H2O(l)
[22.65]
Carbon dioxide is a colorless, odorless gas. It is a minor component of Earth’s atmosphere but a major contributor to the greenhouse effect. • (Section 18.2) Although it is not toxic, high concentrations of CO2 increase respiration rate and can cause suffocation. It is readily liquefied by compression. When cooled at atmospheric pressure, however, CO2 forms a solid rather than liquefying. The solid sublimes at atmospheric pressure at -78 °C. This property makes solid CO2, known as dry ice, valuable as a refrigerant. About half of the CO2 consumed annually is used for refrigeration. The other major use of CO2 is in the production of carbonated beverages. Large quantities are also used to manufacture washing soda (Na2CO3 # 10 H2O), used to precipitate metal ions that interfere with the cleansing action of soap, and baking soda (NaHCO3). Baking soda is so named because this reaction occurs during baking: NaHCO3(s) + H+(aq) ¡ Na +(aq) + CO2(g) + H2O(l)
[22.66]
쑿 FIGURE 22.32 CO2 formation from the reaction between an acid and calcium carbonate in rock.
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The H+(aq) is provided by vinegar, sour milk, or the hydrolysis of certain salts. The bubbles of CO2 that form are trapped in the baking dough, causing it to rise. GIVE IT SOME THOUGHT Yeast are living organisms that make bread rise in the absence of baking soda and acid. What must the yeast be producing to make bread rise?
Carbonic Acid and Carbonates Carbon dioxide is moderately soluble in H2O at atmospheric pressure. The resulting solution is moderately acidic because of the formation of carbonic acid (H2CO3): CO2(aq) + H2O(l) Δ H2CO3(aq)
[22.67]
Carbonic acid is a weak diprotic acid. Its acidic character causes carbonated beverages to have a sharp, slightly acidic taste. Although carbonic acid cannot be isolated, hydrogen carbonates (bicarbonates) and carbonates can be obtained by neutralizing carbonic acid solutions. Partial neutralization produces HCO3 -, and complete neutralization gives CO3 2-. The HCO3 - ion is a stronger base than acid 1Kb = 2.3 * 10 - 8; Ka = 5.6 * 10-112. The carbonate ion is much more strongly basic 1Kb = 1.8 * 10 - 42. The principal carbonate minerals are calcite (CaCO3), magnesite (MgCO3), dolomite [MgCa(CO3)2], and siderite (FeCO3). Calcite is the principal mineral in limestone and the main constituent of marble, chalk, pearls, coral reefs, and the shells of marine animals such as clams and oysters. Although CaCO3 has low solubility in pure water, it dissolves readily in acidic solutions with evolution of CO2: CaCO3(s) + 2 H+(aq) Δ Ca2 + (aq) + H2O(l) + CO2(g)
[22.68]
Because water containing CO2 is slightly acidic (Equation 22.67), CaCO3 dissolves slowly in this medium: CaCO3(s) + H2O(l) + CO2(g) ¡ Ca2+(aq) + 2 HCO3-(aq)
[22.69]
This reaction occurs when surface waters move underground through limestone deposits. It is the principal way Ca2+ enters groundwater, producing “hard water.” If the limestone deposit is deep enough underground, dissolution of the limestone produces a cave. One of the most important reactions of CaCO3 is its decomposition into CaO and CO2 at elevated temperatures (Equation 22.63). About 2 * 1010 kg (20 million tons) of calcium oxide, known as lime or quicklime, is produced in the United States annually. Because calcium oxide reacts with water to form Ca(OH)2, it is an important commercial base. It is also important in making mortar, the mixture of sand, water, and CaO used to bind bricks, blocks, or rocks together. Calcium oxide reacts with water and CO2 to form CaCO3, which binds the sand in the mortar: CaO(s) + H2O(l) Δ Ca2 + (aq) + 2 OH-(aq)
[22.70]
Ca2+(aq) + 2 OH-(aq) + CO2(aq) ¡ CaCO3(s) + H2O(l)
[22.71]
Carbides The binary compounds of carbon with metals, metalloids, and certain nonmetals are called carbides. The more active metals form ionic carbides, and the most common of these contain the acetylide ion (C2 2-). This ion is isoelectronic with N2, and its Lewis structure, 3:C ‚ C:42-, has a carbon–carbon triple bond. The most important ionic carbide is calcium carbide (CaC2), produced by the reduction of CaO with carbon at high temperature: 2 CaO(s) + 5 C(s) ¡ 2 CaC2(s) + CO2(g)
[22.72]
The Other Group 4A Elements: Si, Ge, Sn, and Pb
SECTION 22.10
The carbide ion is a very strong base that reacts with water to form acetylene (H ¬ C ‚ C ¬ H): CaC2(s) + 2 H2O(l) ¡ Ca(OH)2(aq) + C2H2(g)
[22.73]
Calcium carbide is therefore a convenient solid source of acetylene, which is used in welding (Figure 22.13). Interstitial carbides are formed by many transition metals. The carbon atoms occupy open spaces (interstices) between the metal atoms in a manner analogous to the interstitial hydrides (Section 22.2). This process generally makes the metal harder. Tungsten carbide, for example, is very hard and very heat-resistant and, thus, used to make cutting tools. Covalent carbides are formed by boron and silicon. Silicon carbide (SiC), known as Carborundum®, is used as an abrasive and in cutting tools. Almost as hard as diamond, SiC has a diamondlike structure with alternating Si and C atoms.
Other Inorganic Compounds of Carbon Hydrogen cyanide, HCN, is an extremely toxic gas that famously has the odor of bitter almonds. It is produced by the reaction of a cyanide salt, such as NaCN, with an acid. Aqueous solutions of HCN are known as hydrocyanic acid. Neutralization with a base produces cyanide salts, which are used in the manufacture of several plastics, including nylon and Orlon®. The CN - ion forms stable complexes with most transition metals. • (Section 17.5) Carbon disulfide, CS2, is an important industrial solvent for waxes, greases, celluloses, and other nonpolar substances. It is a colorless, volatile liquid (bp 46.3 °C). The vapor is very poisonous and highly flammable. GIVE IT SOME THOUGHT Based on what you know of their physical properties, does CS2 have stronger intermolecular forces than CO2? Explain.
|
22.10 THE OTHER GROUP 4A ELEMENTS: Si, Ge, Sn, AND Pb The trend from nonmetallic to metallic character as we go down a family is strikingly evident in group 4A. Carbon is a nonmetal; silicon and germanium are metalloids; tin and lead are metals. In this section we consider a few general characteristics of group 4A and then look more thoroughly at silicon.
General Characteristics of the Group 4A Elements 2
2
The group 4A elements possess the outer-shell electron configuration ns np . The electronegativities of the elements are generally low (쑼 TABLE 22.8); carbides that formally contain C4- ions are observed only in the case of a few compounds of carbon with very active metals. Formation of 4+ ions by electron loss is not observed for any of these elements; the ionization energies are too high. The +2 oxidation state is found in the TABLE 22.8 • Some Properties of the Group 4A Elements Property Atomic radius (Å) First ionization energy (kJ>mol) Electronegativity X ¬ X single-bond enthalpy (kJ>mol)
C
Si
Ge
Sn
Pb
0.77 1086 2.5 348
1.17 786 1.8 226
1.22 762 1.8 188
1.40 709 1.8 151
1.46 716 1.9 —
4A 6 C 14 Si 32 Ge 50 Sn 82 Pb
949
950
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Chemistry of the Nonmetals
GO FIGURE
What limits the range of temperatures you can use for zone refining of silicon?
chemistry of germanium, tin, and lead, however, and it is the principal oxidation state for lead. The vast majority of the compounds of the group 4A elements are covalently bonded. Carbon, except in highly unusual examples, forms a maximum of four bonds. The other members of the family are able to form more than four bonds. •(Section 8.7) Table 22.8 shows that the strength of a bond between two atoms of a given element decreases as we go down group 4A. Carbon–carbon bonds are quite strong. Carbon, therefore, has a striking ability to form compounds in which carbon atoms are bonded to one another in extended chains and rings, which accounts for the large number of organic compounds that exist. Other elements can form chains and rings, but these bonds are far less important in the chemistries of these other elements. The Si ¬ Si bond strength (226 kJ>mol), for example, is much lower than the Si ¬ O bond strength (386 kJ>mol). As a result, the chemistry of silicon is dominated by the formation of Si ¬ O bonds, and Si ¬ Si bonds play a minor role in silicon chemistry.
Occurrence and Preparation of Silicon Molten section
As heating coil slowly moves down, impurities concentrate in molten section, leaving behind ultrapure Si Silicon rod
Silicon is the second most abundant element, after oxygen, in Earth’s crust. It occurs in SiO2 and in an enormous variety of silicate minerals. The element is obtained by the reduction of molten silicon dioxide with carbon at high temperature: SiO2(l) + 2 C(s) ¡ Si(l) + 2 CO(g)
Elemental silicon has a diamond-type structure. Crystalline silicon is a gray metalliclooking solid that melts at 1410 °C. The element is a semiconductor, as we saw in Chapters 7 and 12, and is used to make solar cells and transistors for computer chips. To be used as a semiconductor, it must be extremely pure, possessing less than 10-7% (1 ppb) impurities. One method of purification is to treat the element with Cl2 to form SiCl4, a volatile liquid that is purified by fractional distillation and then converted back to elemental silicon by reduction with H2: SiCl 4(g) + 2 H2(g) ¡ Si(s) + 4 HCl(g)
Inert atmosphere 쑿 FIGURE 22.33 Zone-refining apparatus for production of ultrapure silicon.
[22.74]
[22.75]
The process known as zone refining can further purify the element (씱 FIGURE 22.33). As a heated coil is passed slowly along a silicon rod, a narrow band of the element is melted. As the molten section is swept slowly along the length of the rod, the impurities concentrate in this section, following it to the end of the rod. The purified top portion of the rod crystallizes as 99.999999999% pure silicon.
Silicates Silicon dioxide and other compounds that contain silicon and oxygen make up over 90% of Earth’s crust. In silicates, a silicon atom is surrounded by four oxygens and silicon is found in its most common oxidation state, +4. The orthosilicate ion, SiO4 4-, is found in very few silicate minerals, but we can view it as the “building block” for many mineral structures. As 씰 FIGURE 22.34 shows, adjacent tetrahedra are linked by a common oxygen atom. Two tetrahedra joined in this way, called the disilicate ion, contain two Si atoms and seven O atoms. Silicon and oxygen are in the +4 and -2 oxidation states, respectively, in all silicates, so the overall charge of any silicate ion must be consistent with these oxidation states. Thus, the charge on Si2O7 is (2)(+4) + (7)(-2) = -6; it is the Si2O7 6- ion. In most silicate minerals, silicate tetrahedra are linked together to form chains, sheets, or three-dimensional structures. We can connect two vertices of each tetrahedron to two other tetrahedra, for example, leading to an infinite chain with an Á O ¬ Si ¬ O ¬ Si Á backbone. As Figure 22.34(b) shows, this chain can be viewed as repeating units of the Si2O6 4 - ion or, in terms of its simplest formula, SiO3 2-. The mineral enstatite (MgSiO3) consists of rows of single-strand silicate chains with Mg 2+ ions between the strands to balance charge. In Figure 22.34(c) each silicate tetrahedron is linked to three others, forming an infinite sheet structure. The simplest formula of this sheet is Si2O5 2-. The mineral talc, also known as talcum powder, has the formula Mg3(Si2O5)2(OH)2 and is based on this sheet structure. The Mg 2+ and OH- ions lie between the silicate sheets. The slippery feel of talcum powder is due to the silicate sheets sliding relative to one another.
SECTION 22.10
The Other Group 4A Elements: Si, Ge, Sn, and Pb
Si2O52⫺ repeating unit
4⫺
Si2O64⫺ repeating unit
O O
Si
951
O
O
Silicate ion
Fragment of silicate chain
Fragment of silicate sheet
(a)
(b)
(c)
쑿 FIGURE 22.34 Silicate chains and sheets.
Many minerals are based on silicates, and many are useful as clays, ceramics, and other materials. Some silicates, however, have harmful effects on human health, the best-known example being asbestos, a general term applied to a group of fibrous silicate minerals. The structure of these minerals is either chains of silicate tetrahedra or sheets formed into rolls. The result is that the minerals have a fibrous character (씰 FIGURE 22.35). Asbestos minerals were once widely used as thermal insulation, especially in high-temperature applications, because of the great chemical stability of the silicate structure. In addition, the fibers can be woven into asbestos cloth, which was used for fireproof curtains and other applications. However, the fibrous structure of asbestos minerals poses a health risk because the fibers readily penetrate soft tissues, such as the lungs, where they can cause diseases, including cancer. The use of asbestos as a common building material has therefore been discontinued. When all four vertices of each SiO4 tetrahedron are linked to other tetrahedra, the structure extends in three dimensions. This linking of the tetrahedra forms quartz (SiO2). Because the structure is locked together in a three-dimensional array much like diamond • (Section 12.7) quartz is harder than strand- or sheet-type silicates.
SAMPLE EXERCISE 22.9
Determining an Empirical Formula
The mineral chrysotile is a noncarcinogenic asbestos mineral that is based on the sheet structure shown in Figure 22.34(c). In addition to silicate tetrahedra, the mineral contains Mg 2+ and OH- ions. Analysis of the mineral shows that there are 1.5 Mg atoms per Si atom. What is the empirical formula for chrysotile? SOLUTION Analyze A mineral is described that has a sheet silicate structure with Mg 2+ and OH- ions to balance charge and 1.5 Mg for each 1 Si. We are asked to write the empirical formula for the mineral. Plan As shown in Figure 22.34(c), the silicate sheet structure is based on the Si2O5 2- ion. We first add Mg 2+ to give the proper Mg : Si ratio. We then add OH- ions to obtain a neutral compound. Solve The observation that the Mg : Si ratio equals 1.5 is consistent with three Mg 2+ ions per Si2O5 2- ion. The addition of three Mg 2+ ions would make Mg 3 (Si2O5)4+. In order to achieve charge balance in the mineral, there must be four OH- ions per Si2O5 2- ion. Thus, the formula of chrysotile is Mg3(Si2O5)(OH)4. Since this is not reducible to a simpler formula, this is the empirical formula.
쑿 FIGURE 22.35 Serpentine asbestos.
952
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Chemistry of the Nonmetals
PRACTICE EXERCISE The cyclosilicate ion consists of three silicate tetrahedra linked together in a ring. The ion contains three Si atoms and nine O atoms. What is the overall charge on the ion? Answer: 6 -
Glass Quartz melts at approximately 1600 °C, forming a tacky liquid. In the course of melting, many silicon–oxygen bonds are broken. When the liquid cools rapidly, silicon–oxygen bonds are re-formed before the atoms are able to arrange themselves in a regular fashion. An amorphous solid, known as quartz glass or silica glass, results. Many substances can be added to SiO2 to cause it to melt at a lower temperature. The common glass used in windows and bottles, known as soda-lime glass, contains CaO and Na2O in addition to SiO2 from sand. The CaO and Na2O are produced by heating two inexpensive chemicals, limestone (CaCO3) and soda ash (Na2CO3), which decompose at high temperatures: CaCO3(s) ¡ CaO(s) + CO2(g)
[22.76]
Na2CO3(s) ¡ Na2O(s) + CO2(g)
[22.77]
Other substances can be added to soda-lime glass to produce color or to change the properties of the glass in various ways. The addition of CoO, for example, produces the deep blue color of “cobalt glass.” Replacing Na2O with K2O results in a harder glass that has a higher melting point. Replacing CaO with PbO results in a denser “lead crystal” glass with a higher refractive index. Lead crystal is used for decorative glassware; the higher refractive index gives this glass a particularly sparkling appearance. Addition of nonmetal oxides, such as B2O3 and P4O10, which form network structures related to the silicates, also changes the properties of the glass. Adding B2O3 creates a “borosilicate” glass with a higher melting point and a greater ability to withstand temperature changes. Such glasses, sold commercially under trade names such as Pyrex® and Kimax®, are used where resistance to thermal shock is important, such as in laboratory glassware or coffeemakers.
Silicones Silicones consist of O ¬ Si ¬ O chains in which the remaining bonding positions on each silicon are occupied by organic groups such as CH3: H3C ...
CH3
H3C
Si O
CH3
H3C
Si O
CH3 ...
Si O
O
Depending on chain length and degree of cross-linking, silicones can be either oils or rubber-like materials. Silicones are nontoxic and have good stability toward heat, light, oxygen, and water. They are used commercially in a wide variety of products, including lubricants, car polishes, sealants, and gaskets. They are also used for waterproofing fabrics. When applied to a fabric, the oxygen atoms form hydrogen bonds with the molecules on the surface of the fabric. The hydrophobic (waterrepelling) organic groups of the silicone are then left pointing away from the surface as a barrier.
GIVE IT SOME THOUGHT Distinguish among the substances silicon, silica, and silicone.
SECTION 22.11
Boron
22.11 | BORON Boron is the only group 3A element that can be considered nonmetallic and thus is our final element in this chapter. The element has an extended network structure with a melting point (2300 °C) that is intermediate between the melting points of carbon (3550 °C) and silicon (1410 °C). The electron configuration of boron is 3He42s22p1. In the family of compounds called boranes, the molecules contain only boron and hydrogen. The simplest borane is BH3. This molecule contains only six valence electrons and is therefore an exception to the octet rule. As a result, BH3 reacts with itself to form diborane (B2H6). This reaction can be viewed as a Lewis acid–base reaction in which one B ¬ H bonding pair of electrons in each BH3 molecule is donated to the other. As a result, diborane is an unusual molecule in which hydrogen atoms form a bridge between two B atoms (씰 FIGURE 22.36). Such hydrogens, called bridging hydrogens, exhibit interesting chemical reactivity, which you may learn about in a more advanced chemistry course. Sharing hydrogen atoms between the two boron atoms compensates somewhat for the deficiency in valence electrons around each boron. Nevertheless, diborane is an extremely reactive molecule, spontaneously flammable in air in a highly exothermic reaction: B2H6(g) + 3 O2(g) ¡ B2O3(s) + 3 H2O(g)
GIVE IT SOME THOUGHT Recall that the hydride ion is H-. What is the oxidation state of boron in sodium borohydride?
The only important oxide of boron is boric oxide (B2O3). This substance is the anhydride of boric acid, which we may write as H3BO3 or B(OH)3. Boric acid is so weak an acid (Ka = 5.8 * 10-10) that solutions of H3BO3 are used as an eyewash. Upon heating, boric acid loses water by a condensation reaction similar to that described for phosphorus in Section 22.8: 4 H3BO3(s) ¡ H2B4O7(s) + 5 H2O(g)
[22.79]
The diprotic acid H2B4O7 is tetraboric acid. The hydrated sodium salt Na2B4O7 # 10 H2O, called borax, occurs in dry lake deposits in California and can also be prepared from other borate minerals. Solutions of borax are alkaline, and the substance is used in various laundry and cleaning products. Putting Concepts Together
The interhalogen compound BrF3 is a volatile, straw-colored liquid. The compound exhibits appreciable electrical conductivity because of autoionization (“solv” refers to BrF3 as the solvent): 2 BrF3(l) Δ BrF2 +(solv) + BrF4 -(solv) (a) What are the molecular structures of the BrF2 + and BrF4 - ions? (b) The electrical conductivity of BrF3 decreases with increasing temperature. Is the autoionization process exothermic or endothermic? (c) One chemical characteristic of BrF3 is that it acts as a Lewis acid toward fluoride ions. What do we expect will happen when KBr is dissolved in BrF3? SOLUTION (a) The BrF2 + ion has 7 + 2(7) - 1 = 20 valence-shell electrons. The Lewis structure for the ion is F
Br
F
⫹
13 Al 31 Ga 49 In 81 Tl
¢H° = -2030 kJ [22.78]
Boron and hydrogen form a series of anions called borane anions. Salts of the borohydride ion (BH4 -) are widely used as reducing agents. For example, sodium borohydride (NaBH4) is a commonly used reducing agent for certain organic compounds.
SAMPLE INTEGRATIVE EXERCISE
3A 5 B
쑿 FIGURE 22.36 The structure of diborane (B2H6).
953
954
CHAPTER 22
Chemistry of the Nonmetals
Because there are four electron domains around the central Br atom, the resulting electron domain geometry is tetrahedral. • (Section 9.2) Because bonding pairs of electrons occupy two of these domains, the molecular geometry is nonlinear:
⫹
⫹
The BrF4 - ion has 7 + 4(7) + 1 = 36 electrons, leading to the Lewis structure ⫺
F F
Br
F F
Because there are six electron domains around the central Br atom in this ion, the geometry is octahedral. The two nonbonding pairs of electrons are located opposite each other on the octahedron, leading to a square-planar molecular geometry:
⫺
⫺
(b) The observation that conductivity decreases as temperature increases indicates that there are fewer ions present in the solution at the higher temperature. Thus, increasing the temperature causes the equilibrium to shift to the left. According to Le Châtelier’s principle, this shift indicates that the reaction is exothermic as it proceeds from left to right. •(Section 15.7) (c) A Lewis acid is an electron-pair acceptor. • (Section 16.11) The fluoride ion has four valence-shell electron pairs and can act as a Lewis base (an electron-pair donor). Thus, we can envision the following reaction occurring: ⫺
F F⫺ ⫹
F F
Br F
F⫺
⫹
BrF3
F
Br
F F
BrF4⫺
CHAPTER SUMMARY AND KEY TERMS INTRODUCTION AND SECTION 22.1 The periodic table is useful for organizing and remembering the descriptive chemistry of the elements. Among elements of a given group, size increases with increasing atomic number, and electronegativity and ionization energy decrease. Nonmetallic character parallels electronegativity, so the most nonmetallic elements are found in the upper right portion of the periodic table. Among the nonmetallic elements, the first member of each group differs dramatically from the other members; it forms a maximum of four bonds to other atoms and exhibits a much greater tendency to form p bonds than the heavier elements in its group. Because O2 and H2O are abundant in our world, we focus on two important and general reaction types as we discuss the nonmetals: oxidation by O2 and proton-transfer reactions involving H2O or aqueous solutions.
Hydrogen has three isotopes: protium (11H), and tritium (31H). Hydrogen is not a member of any particular periodic group, although it is usually placed above lithium. The hydrogen atom can either lose an electron, forming H+, or gain one, forming H- (the hydride ion). Because the H ¬ H bond is relatively strong, H2 is fairly unreactive unless activated by heat or a catalyst. Hydrogen forms a very strong bond to oxygen, so the reactions of H2 with oxygen-containing compounds usually lead to the formation of H2O. Because the bonds in CO and CO2 are even stronger than the O ¬ H bond, the reaction of H2O with carbon or certain organic compounds leads to the formation of H2. The H+(aq) ion is able to oxidize many metals, forming H2(g). The electrolysis of water also forms H2(g). The binary compounds of hydrogen are of three general types: ionic hydrides (formed by active metals), metallic hydrides (formed SECTION
22.2
deuterium (21H),
Key Skills by transition metals), and molecular hydrides (formed by nonmetals). The ionic hydrides contain the H- ion; because this ion is extremely basic, ionic hydrides react with H2O to form H2 and OH- . The noble gases (group 8A) exhibit a very limited chemical reactivity because of the exceptional stability of their electron configurations. The xenon fluorides and oxides and KrF2 are the best-established compounds of the noble gases. The halogens (group 7A) occur as diatomic molecules. All except fluorine exhibit oxidation states varying from -1 to +7. Fluorine is the most electronegative element, so it is restricted to the oxidation states 0 and -1. The oxidizing power of the element (the tendency to form the -1 oxidation state) decreases as we proceed down the group. The hydrogen halides are among the most useful compounds of these elements; these gases dissolve in water to form the hydrohalic acids, such as HCl(aq). Hydrofluoric acid reacts with silica. The interhalogens are compounds formed between two different halogen elements. Chlorine, bromine, and iodine form a series of oxyacids, in which the halogen atom is in a positive oxidation state. These compounds and their associated oxyanions are strong oxidizing agents. SECTIONS 22.3 AND 22.4
Oxygen has two allotropes, O2 and O3 (ozone). Ozone is unstable compared to O2, and it is a stronger oxidizing agent than O2. Most reactions of O2 lead to oxides, compounds in which oxygen is in the -2 oxidation state. The soluble oxides of nonmetals generally produce acidic aqueous solutions; they are called acidic anhydrides or acidic oxides. In contrast, soluble metal oxides produce basic solutions and are called basic anhydrides or basic oxides. Many metal oxides that are insoluble in water dissolve in acid, accompanied by the formation of H2O. Peroxides contain O ¬ O bonds and oxygen in the -1 oxidation state. Peroxides are unstable, decomposing to O2 and oxides. In such reactions peroxides are simultaneously oxidized and reduced, a process called disproportionation. Superoxides contain the O2 - ion in which oxygen is in the - 12 oxidation state. Sulfur is the most important of the other group 6A elements. It has several allotropic forms; the most stable one at room temperature consists of S8 rings. Sulfur forms two oxides, SO2 and SO3, and both are important atmospheric pollutants. Sulfur trioxide is the anhydride of sulfuric acid, the most important sulfur compound and the mostproduced industrial chemical. Sulfuric acid is a strong acid and a good dehydrating agent. Sulfur forms several oxyanions as well, including the SO3 2- (sulfite), SO4 2- (sulfate), and S2O3 2- (thiosulfate) ions. Sulfur is found combined with many metals as a sulfide, in which sulfur is in the -2 oxidation state. These compounds often react with acids to form hydrogen sulfide (H2S), which smells like rotten eggs. SECTIONS 22.5 AND 22.6
Nitrogen is found in the atmosphere as N2 molecules. Molecular nitrogen is chemically very stable because of the strong N ‚ N bond. Molecular nitrogen can be converted into SECTIONS 22.7 AND 22.8
955
ammonia via the Haber process. Once the ammonia is made, it can be converted into a variety of different compounds that exhibit nitrogen oxidation states ranging from -3 to +5. The most important industrial conversion of ammonia is the Ostwald process, in which ammonia is oxidized to nitric acid (HNO3). Nitrogen has three important oxides: nitrous oxide (N2O), nitric oxide (NO), and nitrogen dioxide (NO2). Nitrous acid (HNO2) is a weak acid; its conjugate base is the nitrite ion (NO2 -). Another important nitrogen compound is hydrazine (N2H4). Phosphorus is the most important of the remaining group 5A elements. It occurs in nature as phosphate minerals. Phosphorus has several allotropes, including white phosphorus, which consists of P4 tetrahedra. In reaction with the halogens, phosphorus forms trihalides PX3 and pentahalides PX5. These compounds undergo hydrolysis to produce an oxyacid of phosphorus and HX. Phosphorus forms two oxides, P4O6 and P4O10. Their corresponding acids, phosphorous acid and phosphoric acid, undergo condensation reactions when heated. Phosphorus compounds are important in biochemistry and as fertilizers. The allotropes of carbon include diamond, graphite, fullerenes, carbon nanotubes, and graphene. Amorphous forms of graphite include charcoal and carbon black. Carbon forms two common oxides, CO and CO2. Aqueous solutions of CO2 produce the weak diprotic acid carbonic acid (H2CO3), which is the parent acid of hydrogen carbonate and carbonate salts. Binary compounds of carbon are called carbides. Carbides may be ionic, interstitial, or covalent. Calcium carbide (CaC2) contains the strongly basic acetylide ion (C2 2-), which reacts with water to form acetylene. Other important inorganic carbon compounds include hydrogen cyanide (HCN) and carbon disulfide (CS2). The other group 4A elements show great diversity in physical and chemical properties. Silicon, the second most abundant element, is a semiconductor. It reacts with Cl2 to form SiCl4, a liquid at room temperature, a reaction that is used to help purify silicon from its native minerals. Silicon forms strong Si ¬ O bonds and therefore occurs in a variety of silicate minerals. Silica is SiO2; silicates consist of SiO4 tetrahedra, linked together at their vertices to form chains, sheets, or three-dimensional structures. The most common three-dimensional silicate is quartz (SiO2). Glass is an amorphous (noncrystalline) form of SiO2. Silicones contain O ¬ Si ¬ O chains with organic groups bonded to the Si atoms. Like silicon, germanium is a metalloid; tin and lead are metallic. SECTIONS 22.9 AND 22.10
Boron is the only group 3A element that is a nonmetal. It forms a variety of compounds with hydrogen called boron hydrides, or boranes. Diborane (B2H6) has an unusual structure with two hydrogen atoms that bridge between the two boron atoms. Boranes react with oxygen to form boric oxide (B2O3), in which boron is in the +3 oxidation state. Boric oxide is the anhydride of boric acid (H3BO3). Boric acid readily undergoes condensation reactions. SECTION 22.11
KEY SKILLS • Be able to use periodic trends to explain the basic differences between the elements of a group or period (Section 22.1). • Explain the ways in which the first element in a group differs from subsequent elements in the group (Section 22.1). • Be able to determine electron configurations, oxidation numbers, and molecular shapes of elements and compounds (Sections 22.2–22.11). • Know the sources of the common nonmetals, how they are obtained, and how they are used (Sections 22.2–22.11). • Understand how phosphoric and phosphorous acids undergo condensation reactions (Section 22.8). • Explain how the bonding and structures of silicates relate to their chemical formulas and properties (Section 22.10).
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Chemistry of the Nonmetals
CHAPTER 22
EXERCISES VISUALIZING CONCEPTS 22.1 (a) One of these structures is a stable compound; the other is not. Identify the stable compound, and explain why it is stable. Explain why the other compound is not stable. (b) What is the geometry around the central atoms of the stable compound? [Section 22.1] H
H C
H
H
H Si
C H
H
22.6 Which property of the group 6A elements might be the one depicted in the graph shown here: (a) electronegativity, (b) first ionization energy, (c) density, (d) X ¬ X single-bond enthalpy, (e) electron affinity? Explain your answer. [Sections 22.5 and 22.6]
O
Si H
22.2 (a) Identify the type of chemical reaction represented by the following diagram. (b) Place appropriate charges on the species on both sides of the equation. (c) Write the chemical equation for the reaction. [Section 22.1]
S Se Te Po
⫹
⫹
22.3 Which of the following species (there may be more than one) is> are likely to have the structure shown here: (a) XeF4, (b) BrF4 + , (c) SiF4, (d) TeCl 4, (e) HClO4? (The colors do not reflect atom identities.) [Sections 22.3, 22.4, 22.6, and 22.10]
22.7 The atomic and ionic radii of the first three group 6A elements are Atomic radius (Å)
Ionic radius (Å)
O
O2⫺
0.73
1.40
S
S2⫺
22.4 You have two glass bottles, one containing oxygen and one filled with ozone. How could you determine which one is which? [Section 22.5]
1.03
1.84
22.5 Write the molecular formula and Lewis structure for each of the following oxides of nitrogen: [Section 22.7]
Se
Se2⫺
1.17
1.98
(a) Explain why the atomic radius increases in moving downward in the group. (b) Explain why the ionic radii are larger than the atomic radii. (c) Which of the three anions would you expect to be the strongest base in water? Explain. [Sections 22.5 and 22.6] 22.8 Which property of the third-row nonmetallic elements might be the one depicted in the graph on the next page: (a) first ionization energy, (b) atomic radius, (c) electronegativity, (d) melting point, (e) X ¬ X single-bond enthalpy? Explain both
Exercises your choice and why the other choices would not be correct. [Sections 22.3, 22.4, 22.6, 22.8, and 22.10]
Si P
22.9 Which of the following compounds would you expect to be the most generally reactive, and why? [Section 22.8]
O
O
O
22.10 (a) Draw the Lewis structures for at least four species that have the general formula
S
n
X
Cl Ar
957
Y
where X and Y may be the same or different, and n may have a value from +1 to -2. (b) Which of the compounds is likely to be the strongest Brønsted base? Explain. [Sections 22.1, 22.7, and 22.9]
PERIODIC TRENDS AND CHEMICAL REACTIONS (section 22.1) 22.11 Identify each of the following elements as a metal, nonmetal, or metalloid: (a) phosphorus, (b) strontium, (c) manganese, (d) selenium, (e) sodium, (f) krypton. 22.12 Identify each of the following elements as a metal, nonmetal, or metalloid: (a) gallium, (b) molybdenum, (c) tellurium, (d) arsenic, (e) xenon, (f) ruthenium. 22.13 Consider the elements O, Ba, Co, Be, Br, and Se. From this list select the element that (a) is most electronegative, (b) exhibits a maximum oxidation state of +7, (c) loses an electron most readily, (d) forms p bonds most readily, (e) is a transition metal, (f) is a liquid at room temperature and pressure. 22.14 Consider the elements Li, K, Cl, C, Ne, and Ar. From this list select the element that (a) is most electronegative, (b) has the greatest metallic character, (c) most readily forms a positive ion, (d) has the smallest atomic radius, (e) forms p bonds most readily, (f) has multiple allotropes. 22.15 Explain the following observations: (a) The highest fluoride compound formed by nitrogen is NF3, whereas phosphorus readily forms PF5. (b) Although CO is a well-known compound, SiO does not exist under ordinary conditions. (c) AsH3 is a stronger reducing agent than NH3.
22.16 Explain the following observations: (a) HNO3 is a stronger oxidizing agent than H3PO4. (b) Silicon can form an ion with six fluorine atoms, SiF6 2- , whereas carbon is able to bond to a maximum of four, CF4. (c) There are three compounds formed by carbon and hydrogen that contain two carbon atoms each (C2H2, C2H4, and C2H6), whereas silicon forms only one analogous compound (Si2H6). 22.17 Complete and balance the following equations: (a) NaOCH3(s) + H2O(l) ¡ (b) CuO(s) + HNO3(aq) ¡ ¢
(c) WO3(s) + H2(g) ¡ (d) NH2OH(l) + O2(g) ¡ (e) Al 4C3(s) + H2O(l) ¡ 22.18 Complete and balance the following equations: (a) Mg 3N2(s) + H2O(l) ¡ (b) C3H7OH(l) + O2(g) ¡ ¢
(c) MnO2(s) + C(s) ¡ (d) AlP(s) + H2O(l) ¡ (e) Na2S(s) + HCl(aq) ¡
HYDROGEN, THE NOBLE GASES, AND THE HALOGENS (sections 22.2, 22.3, 22.4) 22.19 (a) Give the names and chemical symbols for the three isotopes of hydrogen. (b) List the isotopes in order of decreasing natural abundance. (c) Which hydrogen isotope is radioactive? (d) Write the nuclear equation for the radioactive decay of this isotope. 22.20 Are the physical properties of H2O different from D2O? Explain.
22.25 Complete and balance the following equations: (a) NaH(s) + H2O(l) ¡ (b) Fe(s) + H2SO4(aq) ¡ (c) H2(g) + Br2(g) ¡ (d) Na(l) + H2(g) ¡ (e) PbO(s) + H2(g) ¡
22.23 Write a balanced equation for the preparation of H2 using (a) Mg and an acid, (b) carbon and steam, (c) methane and steam.
22.26 Write balanced equations for each of the following reactions (some of these are analogous to reactions shown in the chapter). (a) Aluminum metal reacts with acids to form hydrogen gas. (b) Steam reacts with magnesium metal to give magnesium oxide and hydrogen. (c) Manganese(IV) oxide is reduced to manganese(II) oxide by hydrogen gas. (d) Calcium hydride reacts with water to generate hydrogen gas.
22.24 List (a) three commercial means of producing H2, (b) three industrial uses of H2.
22.27 Identify the following hydrides as ionic, metallic, or molecular: (a) BaH2, (b) H2Te, (c) TiH1.7.
22.21 Give a reason why hydrogen might be placed along with the group 1A elements of the periodic table. 22.22 What does hydrogen have in common with the halogens? Explain.
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22.28 Identify the following hydrides as ionic, metallic, or molecular: (a) B2H6, (b) RbH, (c) Th4H1.5.
(c) iodine trichloride, (d) sodium hypochlorite, (e) perchloric acid, (f) xenon tetrafluoride.
22.29 Describe two characteristics of hydrogen that are favorable for its use as a general energy source in vehicles.
22.35 Name the following compounds and assign oxidation states to the halogens in them: (a) Fe(ClO3)3, (b) HClO2, (c) XeF6, (d) BrF5, (e) XeOF4, (f) HIO3.
22.30 The H2> O2 fuel cell converts elemental hydrogen and oxygen into water, producing, theoretically, 1.23 V of energy. What is the most sustainable way to obtain hydrogen to run a large number of fuel cells? Explain. 22.31 Why does xenon form stable compounds with fluorine, whereas argon does not? 22.32 A friend tells you that the “neon” in neon signs is a compound of neon and aluminum. Can your friend be correct? Explain. 22.33 Write the chemical formula for each of the following, and indicate the oxidation state of the halogen or noble-gas atom in each: (a) calcium hypobromite, (b) bromic acid, (c) xenon trioxide, (d) perchlorate ion, (e) iodous acid, (f) iodine pentafluoride. 22.34 Write the chemical formula for each of the following compounds, and indicate the oxidation state of the halogen or noble-gas atom in each: (a) chlorate ion, (b) hydroiodic acid,
22.36 Name the following compounds and assign oxidation states to the halogens in them: (a) KClO3, (b) Ca(IO3)2, (c) AlCl3, (d) HBrO3, (e) H5IO6, (f) XeF4. 22.37 Explain each of the following observations: (a) At room temperature I2 is a solid, Br2 is a liquid, and Cl2 and F2 are both gases. (b) F2 cannot be prepared by electrolytic oxidation of aqueous F - solutions. (c) The boiling point of HF is much higher than those of the other hydrogen halides. (d) The halogens decrease in oxidizing power in the order F2 7 Cl 2 7 Br2 7 I2. 22.38 Explain the following observations: (a) For a given oxidation state, the acid strength of the oxyacid in aqueous solution decreases in the order chlorine 7 bromine 7 iodine. (b) Hydrofluoric acid cannot be stored in glass bottles. (c) HI cannot be prepared by treating NaI with sulfuric acid. (d) The interhalogen ICl3 is known, but BrCl3 is not.
OXYGEN AND THE OTHER GROUP 6A ELEMENTS (sections 22.5 and 22.6) 22.39 Write balanced equations for each of the following reactions. (a) When mercury(II) oxide is heated, it decomposes to form O2 and mercury metal. (b) When copper(II) nitrate is heated strongly, it decomposes to form copper(II) oxide, nitrogen dioxide, and oxygen. (c) Lead(II) sulfide, PbS(s), reacts with ozone to form PbSO4(s) and O2(g). (d) When heated in air, ZnS(s) is converted to ZnO. (e) Potassium peroxide reacts with CO2(g) to give potassium carbonate and O2. (f) Oxygen is converted to ozone in the upper atmosphere. 22.40 Complete and balance the following equations: (a) CaO(s) + H2O(l) ¡ (b) Al 2O3(s) + H+(aq) ¡ (c) Na2O2(s) + H2O(l) ¡ (d) N2O3(g) + H2O(l) ¡ (e) KO2(s) + H2O(l) ¡ (f) NO(g) + O3(g) ¡ 22.41 Predict whether each of the following oxides is acidic, basic, amphoteric, or neutral: (a) NO2, (b) CO2, (c) Al2O3, (d) CaO. 22.42 Select the more acidic member of each of the following pairs: (a) Mn2O7 and MnO2, (b) SnO and SnO2, (c) SO2 and SO3, (d) SiO2 and SO2, (e) Ga2O3 and In2O3, (f) SO2 and SeO2. 22.43 Write the chemical formula for each of the following compounds, and indicate the oxidation state of the group 6A element in each: (a) selenous acid, (b) potassium hydrogen sulfite, (c) hydrogen telluride, (d) carbon disulfide, (e) calcium sulfate, (f) cadmium sulfide, (g) zinc telluride. 22.44 Write the chemical formula for each of the following compounds, and indicate the oxidation state of the group 6A element in each: (a) sulfur tetrachloride, (b) selenium triox-
ide, (c) sodium thiosulfate, (d) hydrogen sulfide, (e) sulfuric acid, (f) sulfur dioxide, (g) mercury telluride. 22.45 In aqueous solution, hydrogen sulfide reduces (a) Fe 3+ to Fe 2+, (b) Br2 to Br - , (c) MnO4 - to Mn2+ , (d) HNO3 to NO2. In all cases, under appropriate conditions, the product is elemental sulfur. Write a balanced net ionic equation for each reaction. 22.46 An aqueous solution of SO2 reduces (a) aqueous KMnO4 to MnSO4(aq), (b) acidic aqueous K2Cr2O7 to aqueous Cr 3+ , (c) aqueous Hg2(NO3)2 to mercury metal. Write balanced equations for these reactions. 22.47 Write the Lewis structure for each of the following species, and indicate the structure of each: (a) SeO3 2- ; (b) S2Cl2; (c) chlorosulfonic acid, HSO3Cl (chlorine is bonded to sulfur). 22.48 The SF5 - ion is formed when SF4(g) reacts with fluoride salts containing large cations, such as CsF(s). Draw the Lewis structures for SF4 and SF5 - , and predict the molecular structure of each. 22.49 Write a balanced equation for each of the following reactions: (a) Sulfur dioxide reacts with water. (b) Solid zinc sulfide reacts with hydrochloric acid. (c) Elemental sulfur reacts with sulfite ion to form thiosulfate. (d) Sulfur trioxide is dissolved in sulfuric acid. 22.50 Write a balanced equation for each of the following reactions. (You may have to guess at one or more of the reaction products, but you should be able to make a reasonable guess, based on your study of this chapter.) (a) Hydrogen selenide can be prepared by reaction of an aqueous acid solution on aluminum selenide. (b) Sodium thiosulfate is used to remove excess Cl2 from chlorine-bleached fabrics. The thiosulfate ion forms SO4 2and elemental sulfur, while Cl2 is reduced to Cl-.
NITROGEN AND THE OTHER GROUP 5A ELEMENTS (sections 22.7 and 22.8) 22.51 Write the chemical formula for each of the following compounds, and indicate the oxidation state of nitrogen in each: (a) sodium nitrite, (b) ammonia, (c) nitrous oxide, (d) sodium cyanide, (e) nitric acid, (f) nitrogen dioxide, (g) nitrogen, (h) boron nitride.
22.52 Write the chemical formula for each of the following compounds, and indicate the oxidation state of nitrogen in each: (a) nitric oxide, (b) hydrazine, (c) potassium cyanide, (d) sodium nitrite, (e) ammonium chloride, (f) lithium nitride.
Exercises
959
22.53 Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the nitrogen: (a) HNO2, (b) N3 - , (c) N2H5 + , (d) NO3 - .
(a) phosphorous acid, (b) pyrophosphoric acid, (c) antimony trichloride, (d) magnesium arsenide, (e) diphosphorus pentoxide, (f) sodium phosphate.
22.54 Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the nitrogen: (a) NH4 + , (b) NO2 - , (c) N2O, (d) NO2.
22.60 Write a chemical formula for each compound or ion, and indicate the oxidation state of the group 5A element in each formula: (a) phosphate ion, (b) arsenous acid, (c) antimony(III) sulfide, (d) calcium dihydrogen phosphate, (e) potassium phosphide, (f) gallium arsenide.
22.55 Complete and balance the following equations: (a) Mg 3N2(s) + H2O(l) ¡ (b) NO(g) + O2(g) ¡ (c) N2O5(g) + H2O(l) ¡ (d) NH3(aq) + H+(aq) ¡ (e) N2H4(l) + O2(g) ¡ Which ones of these are redox reactions? 22.56 Write a balanced net ionic equation for each of the following reactions: (a) Dilute nitric acid reacts with zinc metal with formation of nitrous oxide. (b) Concentrated nitric acid reacts with sulfur with formation of nitrogen dioxide. (c) Concentrated nitric acid oxidizes sulfur dioxide with formation of nitric oxide. (d) Hydrazine is burned in excess fluorine gas, forming NF3. (e) Hydrazine reduces CrO4 2- to Cr(OH)4 - in base (hydrazine is oxidized to N2). 22.57 Write complete balanced half-reactions for (a) oxidation of nitrous acid to nitrate ion in acidic solution, (b) oxidation of N2 to N2O in acidic solution. 22.58 Write complete balanced half-reactions for (a) reduction of nitrate ion to NO in acidic solution, (b) oxidation of HNO2 to NO2 in acidic solution. 22.59 Write a molecular formula for each compound, and indicate the oxidation state of the group 5A element in each formula:
22.61 Account for the following observations: (a) Phosphorus forms a pentachloride, but nitrogen does not. (b) H3PO2 is a monoprotic acid. (c) Phosphonium salts, such as PH4Cl, can be formed under anhydrous conditions, but they can’t be made in aqueous solution. (d) White phosphorus is more reactive than red phosphorus. 22.62 Account for the following observations: (a) H3PO3 is a diprotic acid. (b) Nitric acid is a strong acid, whereas phosphoric acid is weak. (c) Phosphate rock is ineffective as a phosphate fertilizer. (d) Phosphorus does not exist at room temperature as diatomic molecules, but nitrogen does. (e) Solutions of Na3PO4 are quite basic. 22.63 Write a balanced equation for each of the following reactions: (a) preparation of white phosphorus from calcium phosphate, (b) hydrolysis of PBr3, (c) reduction of PBr3 to P4 in the gas phase, using H2. 22.64 Write a balanced equation for each of the following reactions: (a) hydrolysis of PCl5, (b) dehydration of phosphoric acid (also called orthophosphoric acid) to form pyrophosphoric acid, (c) reaction of P4O10 with water.
CARBON, THE OTHER GROUP 4A ELEMENTS, AND BORON (sections 22.9, 22.10, 22.11) 22.65 Give the chemical formula for (a) hydrocyanic acid, (b) nickel tetracarbonyl, (c) barium bicarbonate, (d) calcium acetylide (e) potassium carbonate. 22.66 Give the chemical formula for (a) carbonic acid, (b) sodium cyanide, (c) potassium hydrogen carbonate, (d) acetylene, (e) iron pentacarbonyl. 22.67 Complete and balance the following equations: (a) (b) (c) (d) (e)
¢
ZnCO3(s) ¡ BaC2(s) + H2O(l) ¡ C2H2(g) + O2(g) ¡ CS2(g) + O2(g) ¡ Ca(CN)2(s) + HBr(aq) ¡
22.68 Complete and balance the following equations: (a) CO2(g) + OH-(aq) ¡ (b) NaHCO3(s) + H+(aq) ¡ ¢
(c) CaO(s) + C(s) ¡ ¢
(d) C(s) + H2O(g) ¡ (e) CuO(s) + CO(g) ¡ 22.69 Write a balanced equation for each of the following reactions: (a) Hydrogen cyanide is formed commercially by passing a mixture of methane, ammonia, and air over a catalyst at 800 °C. Water is a by-product of the reaction. (b) Baking soda reacts with acids to produce carbon dioxide gas. (c) When barium carbonate reacts in air with sulfur dioxide, barium sulfate and carbon dioxide form.
22.70 Write a balanced equation for each of the following reactions: (a) Burning magnesium metal in a carbon dioxide atmosphere reduces the CO2 to carbon. (b) In photosynthesis, solar energy is used to produce glucose (C6H12O6) and O2 from carbon dioxide and water. (c) When carbonate salts dissolve in water, they produce basic solutions. 22.71 Write the formulas for the following compounds, and indicate the oxidation state of the group 4A element or of boron in each: (a) boric acid, (b) silicon tetrabromide, (c) lead(II) chloride, (d) sodium tetraborate decahydrate (borax), (e) boric oxide, (f) germanium dioxide. 22.72 Write the formulas for the following compounds, and indicate the oxidation state of the group 4A element or of boron in each: (a) silicon dioxide, (b) germanium tetrachloride, (c) sodium borohydride, (d) stannous chloride, (e) diborane, (f) boron trichloride. 22.73 Select the member of group 4A that best fits each description: (a) has the lowest first ionization energy, (b) is found in oxidation states ranging from -4 to +4, (c) is most abundant in Earth’s crust. 22.74 Select the member of group 4A that best fits each description: (a) forms chains to the greatest extent, (b) forms the most basic oxide, (c) is a metalloid that can form 2+ ions. 22.75 (a) What is the characteristic geometry about silicon in all silicate minerals? (b) Metasilicic acid has the empirical formula H2SiO3. Which of the structures shown in Figure 22.34 would you expect metasilicic acid to have?
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CHAPTER 22
Chemistry of the Nonmetals
22.76 Speculate as to why carbon forms carbonate rather than silicate analogs. 22.77 (a) How does the structure of diborane (B2H6) differ from that of ethane (C2H6)? (b) Explain why diborane adopts the geometry that it does. (c) What is the significance of the statement that the hydrogen atoms in diborane are described as “hydridic”?
22.78 Write a balanced equation for each of the following reactions: (a) Diborane reacts with water to form boric acid and molecular hydrogen. (b) Upon heating, boric acid undergoes a condensation reaction to form tetraboric acid. (c) Boron oxide dissolves in water to give a solution of boric acid.
ADDITIONAL EXERCISES 22.79 In your own words, define the following terms: (a) allotrope, (b) disproportionation, (c) interhalogen, (d) acidic anhydride, (e) condensation reaction, (f) protium. 22.80 Although the ClO4 - and IO4 - ions have been known for a long time, BrO4 - was not synthesized until 1965. The ion was synthesized by oxidizing the bromate ion with xenon difluoride, producing xenon, hydrofluoric acid, and the perbromate ion. (a) Write the balanced equation for this reaction. (b) What are the oxidation states of Br in the Br-containing species in this reaction? 22.81 Write a balanced equation for the reaction of each of the following compounds with water: (a) SO2(g), (b) Cl2O7(g), (c) Na2O2(s), (d) BaC2(s), (e) RbO2(s), (f) Mg3N2(s), (g) NaH(s). 22.82 What is the anhydride for each of the following acids: (a) H2SO4, (b) HClO3, (c) HNO2, (d) H2CO3, (e) H3PO4? 22.83 Explain why SO2 can be used as a reducing agent but SO3 cannot. 22.84 A sulfuric acid plant produces a considerable amount of heat. This heat is used to generate electricity, which helps reduce operating costs. The synthesis of H2SO4 consists of three main chemical processes: (1) oxidation of S to SO2, (2) oxidation of SO2 to SO3, (3) the dissolving of SO3 in H2SO4 and its reaction with water to form H2SO4. If the third process produces 130 kJ>mol, how much heat is produced in preparing a mole of H2SO4 from a mole of S? How much heat is produced in preparing 5000 pounds of H2SO4?
22.85 (a) What is the oxidation state of P in PO43- and of N in NO3 - ? (b) Why doesn’t N form a stable NO43- ion analogous to P? 22.86 (a) The P4, P4O6, and P4O10 molecules have a common structural feature of four P atoms arranged in a tetrahedron (Figures 22.27 and 22.28). Does this mean that the bonding between the P atoms is the same in all these cases? Explain. (b) Sodium trimetaphosphate (Na3P3O9) and sodium tetrametaphosphate (Na4P4O12) are used as water-softening agents. They contain cyclic P3O9 3- and P4O12 4 - ions, respectively. Propose reasonable structures for these ions. 22.87 Ultrapure germanium, like silicon, is used in semiconductors. Germanium of “ordinary” purity is prepared by the hightemperature reduction of GeO2 with carbon. The Ge is converted to GeCl4 by treatment with Cl2 and then purified by distillation; GeCl4 is then hydrolyzed in water to GeO2 and reduced to the elemental form with H2. The element is then zone refined. Write a balanced chemical equation for each of the chemical transformations in the course of forming ultrapure Ge from GeO2. 22.88 Hydrogen peroxide is capable of oxidizing (a) hydrazine to N2 and H2O, (b) SO2 to SO42- , (c) NO2 - to NO3 - , (d) H2S(g) to S(s), (e) Fe 2+ to Fe 3+ . Write a balanced net ionic equation for each of these redox reactions.
INTEGRATIVE EXERCISES [22.89] (a) How many grams of H2 can be stored in 100.0 kg of the alloy FeTi if the hydride FeTiH2 is formed? (b) What volume does this quantity of H2 occupy at STP? (c) If this quantity of hydrogen was combusted in air to produce liquid water, how much energy could be produced? [22.90] Using the thermochemical data in Table 22.1 and Appendix C, calculate the average Xe ¬ F bond enthalpies in XeF2, XeF4, and XeF6, respectively. What is the significance of the trend in these quantities? 22.91 Hydrogen gas has a higher fuel value than natural gas on a mass basis but not on a volume basis. Thus, hydrogen is not competitive with natural gas as a fuel transported long distances through pipelines. Calculate the heats of combustion of H2 and CH4 (the principal component of natural gas) (a) per mole of each, (b) per gram of each, (c) per cubic meter of each at STP. Assume H2O(l) as a product. 22.92 The solubility of Cl2 in 100 g of water at STP is 310 cm3. Assume that this quantity of Cl2 is dissolved and equilibrated as follows: Cl 2(aq) + H2O Δ Cl-(aq) + HClO(aq) + H+(aq)
(a) If the equilibrium constant for this reaction is 4.7 * 10 - 4, calculate the equilibrium concentration of HClO formed. (b) What is the pH of the final solution? [22.93] When ammonium perchlorate decomposes thermally, the products of the reaction are N2(g), O2(g), H2O(g), and HCl(g). (a) Write a balanced equation for the reaction. [Hint: You might find it easier to use fractional coefficients for the products.] (b) Calculate the enthalpy change in the reaction per mole of NH4ClO4. The standard enthalpy of formation of NH4ClO4(s) is -295.8 kJ. (c) When NH4ClO4(s) is employed in solid-fuel booster rockets, it is packed with powdered aluminum. Given the high temperature needed for NH4ClO4(s) decomposition and what the products of the reaction are, what role does the aluminum play? (d) Calculate the volume of all the gases that would be produced at STP, assuming complete reaction of one pound of ammonium perchlorate. 22.94 The dissolved oxygen present in any highly pressurized, hightemperature steam boiler can be extremely corrosive to its metal parts. Hydrazine, which is completely miscible with water, can be added to remove oxygen by reacting with it to form nitrogen and water. (a) Write the balanced equation for
Integrative Exercises the reaction between gaseous hydrazine and oxygen. (b) Calculate the enthalpy change accompanying this reaction. (c) Oxygen in air dissolves in water to the extent of 9.1 ppm at 20 °C at sea level. How many grams of hydrazine are required to react with all the oxygen in 3.0 * 104 L (the volume of a small swimming pool) under these conditions? 22.95 One method proposed for removing SO2 from the flue gases of power plants involves reaction with aqueous H2S. Elemental sulfur is the product. (a) Write a balanced chemical equation for the reaction. (b) What volume of H2S at 27 °C and 760 torr would be required to remove the SO2 formed by burning 2.0 tons of coal containing 3.5% S by mass? (c) What mass of elemental sulfur is produced? Assume that all reactions are 100% efficient. 22.96 The maximum allowable concentration of H2S(g) in air is 20 mg per kilogram of air (20 ppm by mass). How many grams of FeS would be required to react with hydrochloric acid to produce this concentration at 1.00 atm and 25 °C in an average room measuring 12 ft * 20 ft * 8 ft? (Under these conditions, the average molar mass of air is 29.0 g>mol.) 22.97 The standard heats of formation of H2O(g), H2S(g), H2Se(g), and H2Te(g) are -241.8, -20.17, +29.7, and +99.6 kJ>mol, respectively. The enthalpies necessary to convert the elements in their standard states to one mole of gaseous atoms are 248, 277, 227, and 197 kJ>mol of atoms for O, S, Se, and Te, respectively. The enthalpy for dissociation of H2 is 436 kJ>mol. Calculate the average H ¬ O, H ¬ S, H ¬ Se, and H ¬ Te bond enthalpies, and comment on their trend. 22.98 Manganese silicide has the empirical formula MnSi and melts at 1280 °C. It is insoluble in water but does dissolve in aqueous HF. (a) What type of compound do you expect MnSi to be: metallic, molecular, covalent-network, or ionic? (b) Write a likely balanced chemical equation for the reaction of MnSi with concentrated aqueous HF. [22.99] Chemists tried for a long time to make molecular compounds containing silicon–silicon double bonds; they finally succeed in 1981. The trick is having large, bulky R groups on the silicon atoms to make R2Si “ SiR2 compounds. What experiments could you do to prove that a new compound has a silicon–silicon double bond rather than a silicon–silicon single bond? 22.100 Hydrazine has been employed as a reducing agent for metals. Using standard reduction potentials, predict whether the following metals can be reduced to the metallic state by hydrazine under standard conditions in acidic solution: (a) Fe 2+, (b) Sn2+ , (c) Cu2+ , (d) Ag + , (e) Cr 3+ , (f) Co3+ . 22.101 Both dimethylhydrazine, (CH3)2NNH2, and methylhydrazine, CH3NHNH2, have been used as rocket fuels. When dinitrogen tetroxide (N2O4) is used as the oxidizer, the products are H2O, CO2, and N2. If the thrust of the rocket depends on the
961
volume of the products produced, which of the substituted hydrazines produces a greater thrust per gram total mass of oxidizer plus fuel? [Assume that both fuels generate the same temperature and that H2O(g) is formed.] 22.102 Carbon forms an unusual unstable oxide of formula C3O2, called carbon suboxide. Carbon suboxide is made by using P2O5 to dehydrate the dicarboxylic acid called malonic acid, which has the formula HOOC ¬ CH2 ¬ COOH. (a) Write a balanced reaction for the production of carbon suboxide from malonic acid. (b) How many grams of carbon suboxide could be made from 20.00 g of malonic acid? (c) Suggest a Lewis structure for C3O2. (Hint: The Lewis structure of malonic acid suggests which atoms are connected to which.) (d) By using the information in Table 8.5, predict the C ¬ C and C ¬ O bond lengths in C3O2. (e) Sketch the Lewis structure of a product that could result by the addition of 2 mol of H2 to 1 mol of C3O2. 22.103 Borazine, (BH)3(NH)3, is an analog of C6H6, benzene. It can be prepared from the reaction of diborane with ammonia, with hydrogen as another product; or from lithium borohydride and ammonium chloride, with lithium chloride and hydrogen as the other products. (a) Write balanced chemical equations for the production of borazine using both synthetic methods. (b) Draw the Lewis dot structure of borazine. (c) How many grams of borazine can be prepared from 2.00 L of ammonia at STP, assuming diborane is in excess? 22.104 Throughout history, arsenic(III) oxide, known simply to the general public as “arsenic,” has been a poison favored by murderers: It is tasteless, colorless, can be easily added to food or drink, produces symptoms that are similar to several diseases, and until the mid-1800s, was undetectable in the body. James Marsh developed the famous “Marsh test” for arsenic that was instrumental in convicting the murderer in a famous poisoning case in France in 1840. The Marsh test relies on the reaction of arsenic(III) oxide in a sample with elemental zinc and sulfuric acid to produce arsine (an analog of ammonia), zinc sulfate, and water. Upon igniting the final mixture, arsine is oxidized to elemental arsenic, and if captured on a ceramic bowl, a characteristic silvery-black powder would appear. (a) Write the balanced chemical equations of the Marsh test. (b) Antimony is the only potential interferent, as it reacts similarly to arsenic and produces a similar silvery-black film; however, antimony does not dissolve in a solution of sodium hypochlorite, but arsenic does. Therefore, the completion of the Marsh test is to add a solution of sodium hypochlorite to the elemental arsenic and see if the silvery-black film dissolves. Write the balanced chemical equation for this final reaction. (c) Today, commercial kits for arsenic testing rely on a different reaction. The sample is reacted with hydrogen sulfide in the presence of hydrochloric acid; if arsenic is present, As2S3, which is a bright yellow precipitate, forms. Write the balanced equation for this reaction.
WHAT’S AHEAD 23.1 THE TRANSITION METALS We examine the physical properties, electron configurations, oxidation states, and magnetic properties of the transition metals.
23.2 TRANSITION-METAL COMPLEXES We introduce the concepts of metal complexes and ligands and provide a brief history of the development of coordination chemistry.
23.4 NOMENCLATURE AND ISOMERISM IN COORDINATION CHEMISTRY We introduce the nomenclature used for coordination compounds. We see that coordination compounds exhibit isomerism, in which two compounds have the same composition but different structures, and then look at two types: structural isomers and stereoisomers.
23.3 COMMON LIGANDS IN COORDINATION CHEMISTRY We examine some common geometries found in coordination complexes and how the geometries relate to coordination numbers.
23
THIS STAINED GLASS WINDOW, representing chemistry, is in the Kalamazoo Public Library, Kalamazoo, Michigan.
23.5 COLOR AND MAGNETISM IN
23.6 CRYSTAL-FIELD THEORY
COORDINATION CHEMISTRY
We explore how crystal-field theory allows us to explain some of the interesting spectral and magnetic properties of coordination compounds.
We discuss color and magnetism in coordination compounds, emphasizing the visible portion of the electromagnetic spectrum and the notion of complementary colors. We then see that many transition-metal complexes are paramagnetic because they contain unpaired electrons.
TRANSITION METALS AND COORDINATION CHEMISTRY THE COLORS OF OUR WORLD ARE beautiful,
but to a chemist they are also informative—providing insights into the structure and bonding of matter. Compounds of the transition metals constitute an important group of colored substances. Some of them are used in paint pigments; others produce the colors in glass and precious gems. For example, the colors in the stained-glass artwork shown in the chapter-opening photograph are due mainly to transition-metal compounds. Why do these compounds have color, and why do the colors change as the ions or molecules bonded to the metal change? The chemistry we explore in this chapter will help us to answer these questions. In earlier chapters we saw that metal ions can function as Lewis acids, forming covalent bonds with molecules and ions functioning as Lewis bases. • (Section 16.11) We have encountered many ions and compounds that result from such interactions, such as 3Fe(H2O)643 + and 3Ag(NH3)24 + in Sections 16.11 and 17.5 and hemoglobin in 963
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Section 13.6. In this chapter, we focus on the rich and important chemistry associated with such complex assemblies of metal ions surrounded by molecules and ions. Metal compounds of this kind are called coordination compounds, and the branch of chemistry that focuses on them is called coordination chemistry.
23.1 | THE TRANSITION METALS
8B
3B 3 21 Sc
4B 4 22 Ti
5B 5 23 V
6B 6 24 Cr
7B 7 25 Mn
26 Fe
9 27 Co
39 Y
40 Zr
41 Nb
42 Mo
43 Tc
44 Ru
45 Rh
71 Lu
72 Hf
73 Ta
74 W
75 Re
76 Os
77 Ir
8
쑿 FIGURE 23.1 The position of the transition metals in the periodic table. They are the B groups in periods 4, 5, and 6.
The part of the periodic table in which the d orbitals are being filled as we move left to right across a row is the home of the transition metals (씱 FIGURE 23.1). • (Section 6.8) With some exceptions (e.g., platinum, gold), metallic elements are found in nature as solid inorganic compounds called minerals. Notice from 쑼 TABLE 23.1 that minerals are identified by common names rather than chemical names. These common names are usually based on the location where a mineral was discovered, the person who discovered it, or some physical characteristic such as color. 1B 2B Most transition metals in minerals have oxidation states of 10 11 12 +1 , +2, or +3. To obtain a pure metal from its mineral, various 30 29 28 chemical processes must be performed to reduce the metal to the Cu Zn Ni 0 oxidation state. Metallurgy is the science and technology of ex48 47 46 tracting metals from their natural sources and preparing them Pd Ag Cd for practical use. It usually involves several steps: (1) mining, that 80 79 78 is, removing the relevant ore (a mixture of minerals) from the Au Hg Pt ground, (2) concentrating the ore or otherwise preparing it for further treatment, (3) reducing the ore to obtain the free metal, (4) purifying the metal, and (5) mixing it with other elements to modify its properties. This last process produces an alloy, a metallic material composed of two or more elements. • (Section 12.3)
Physical Properties Some physical properties of the period 4 (also known as “first-row”) transition metals are listed in 씰 TABLE 23.2. The properties of the heavier transition metals vary similarly across periods 5 and 6. 씰 FIGURE 23.2 shows the atomic radius observed in close-packed metallic structures as a function of group number.* The trends seen in the graph are a result of two competing forces. On the one hand, increasing effective nuclear charge favors a decrease TABLE 23.1 • Principal Mineral Sources of Some Transition Metals Metal
Mineral
Mineral Composition
Chromium Copper
Chromite Chalcocite Chalcopyrite Malachite Hematite Magnetite Pyrolusite Cinnabar Molybdenite Rutile Ilmenite Sphalerite
FeCr2O4 Cu2S CuFeS2 Cu2CO3(OH)2 Fe2O3 Fe3O4 MnO2 HgS MoS2 TiO2 FeTiO3 ZnS
Iron Manganese Mercury Molybdenum Titanium Zinc
*Note that the radii defined in this way, often referred to as metallic radii, differ somewhat from the bonding atomic radii defined in Section 7.3.
SECTION 23.1
965
The Transition Metals
TABLE 23.2 • Properties of the Period 4 Transition Metals Group
3B
4B
5B
6B
7B
Element:
Sc
Ti
V
Cr
Mn
Fe
Co
3d14s2
3d24s2
3d34s2
3d54s1
3d54s2
3d64s2
First ionization energy (kJ> mol)
631
658
650
653
717
Radius in metallic substances (Å)
1.64
1.47
1.35
1.29
Density (g> cm3)
3.0
4.5
6.1
1541
1660
1917
Melting point (°C)
2B
Ni
Cu
Zn
3d74s2
3d84s2
3d104s1
3d104s2
759
758
737
745
906
1.37
1.26
1.25
1.25
1.28
1.37
7.9
7.2
7.9
8.7
8.9
8.9
7.1
1857
1244
1537
1494
1455
1084
420
in radius as we move left to right across each period. •(Section 7.2) On the other hand, the metallic bonding strength increases until we reach the middle of each period and then decreases as we fill antibonding orbitals. •(Section 12.4) As a general rule, a bond shortens as it becomes stronger. •(Section 8.8) For groups 3B through 6B, these two effects work cooperatively and the result is a marked decrease in radius. In elements to the right of group 6B, the two effects counteract each other, reducing the decrease and eventually leading to an increase in radius. GIVE IT SOME THOUGHT Which element has the largest bonding atomic radius: Sc, Fe, or Au?
1.9 Period 4 (Sc–Zn) Period 5 (Y–Cd) Period 6 (Lu–Hg)
1.8
Atomic radius (Å)
Ground state electron configuration
1B
8B
1.7 1.6 1.5 1.4
In general, atomic radii increase as we move down in a family in 1.3 the periodic table because of the increasing principal quantum number of the outer-shell electrons. • (Section 7.3) Note in Figure 1.2 23.2, however, that once we move beyond the group 3B elements, the 3B 4B period 5 and period 6 transition elements in a given family have virtually the same radii. In group 5B, for example, tantalum in period 6 has virtually the same radius as niobium in period 5. This interesting 쑿 FIGURE 23.2 group number. and important effect has its origin in the lanthanide series, elements 57 through 70. The filling of 4f orbitals through the lanthanide elements • (Figure 6.31) causes a steady increase in the effective nuclear charge, producing a size decrease, called the lanthanide contraction, that just offsets the increase we expect as we go from period 5 transition metals to period 6. Thus, the period 5 and period 6 transition metals in each group have about the same radii all the way across the periodic table. Consequently, the period 5 and period 6 transition metals in a given group have similar chemical properties. For example, the chemical properties of the group 4B metals zirconium (period 5) and hafnium (period 6) are remarkably similar. These two metals always occur together in nature, and they are very difficult to separate.
Electron Configurations and Oxidation States Transition metals owe their location in the periodic table to the filling of the d subshells, as you saw in Figure 6.31. Many of the chemical and physical properties of transition metals result from the unique characteristics of the d orbitals. For a given transitionmetal atom, the valence (n - 1)d orbitals are smaller than the corresponding valence ns and np orbitals. In quantum mechanical terms, the (n - 1)d orbital wave functions drop off more rapidly as we move away from the nucleus than do the ns and np orbital wave functions. This characteristic feature of the d orbitals limits their interaction with orbitals on neighboring atoms, but not so much that they are insensitive to
5B
6B
7B Group
8B
1B
2B
Radii of transition metals as a function of
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CHAPTER 23
Transition Metals and Coordination Chemistry
GO FIGURE
In which transition-metal ion of this group are the 3d orbitals completely filled?
씰 FIGURE 23.3 Aqueous solutions of transition metal ions. Left to right: Co2+ , Ni2+ , Cu2+ , and Zn2+ . The counterion is nitrate in all cases.
surrounding atoms. As a result, electrons in these orbitals behave sometimes like valence electrons and sometimes like core electrons. The details depend on location in the periodic table and the atom’s environment. When transition metals are oxidized, they lose their outer s electrons before they lose electrons from the d subshell. • (Section 7.4) The electron configuration of Fe is 3Ar43d64s2, for example, whereas that of Fe 2 + is 3Ar43d6. Formation of Fe 3 + requires loss of one 3d electron, giving 3Ar43d5. Most transition-metal ions contain partially occupied d subshells, which are responsible in large part for three characteristics:
GO FIGURE
Why does the maximum oxidation state increase linearly from Sc through Mn? Most frequently seen Less common 3B 4B
5B 6B 7B
Sc
V
8B
1B 2B
⫹8 ⫹6 ⫹4 ⫹2 0
Ti
Cr Mn Fe Co Ni Cu Zn
쑿 FIGURE 23.4 Nonzero oxidation states of the period 4 transition metals.
1. Transition metals often have more than one stable oxidation state. 2. Many transition-metal compounds are colored, as shown in 쑿 FIGURE 23.3. 3. Transition metals and their compounds often exhibit magnetic properties. 씱 FIGURE 23.4 shows the common nonzero oxidation states for the period 4 transition metals. The +2 oxidation state, which is common for most transition metals, is due to the loss of the two outer 4s electrons. This oxidation state is found for all these elements except Sc, where the 3+ ion with an [Ar] configuration is particularly stable. Oxidation states above +2 are due to successive losses of 3d electrons. From Sc through Mn the maximum oxidation state increases from +3 to +7, equaling in each case the total number of 4s plus 3d electrons in the atom. Thus, manganese has a maximum oxidation state of 2 + 5 = +7. As we move to the right beyond Mn in Figure 23.4, the maximum oxidation state decreases. This decrease is due in part to the attraction of d orbital electrons to the nucleus, which increases faster than the attraction of the s orbital electrons to the nucleus as we move left to right across the periodic table. In other words, in each period the d electrons become more corelike as the atomic number increases. By the time we get to zinc, it is not possible to remove electrons from the 3d orbitals through chemical oxidation.
SECTION 23.1
In the transition metals of periods 5 and 6, the increased size of the 4d and 5d orbitals makes it possible to attain maximum oxidation states as high as +8, which is achieved in RuO4 and OsO4. In general, the maximum oxidation states are found only when the metals are combined with the most electronegative elements, especially O, F, and in some cases Cl.
The Transition Metals
967
GO FIGURE
Describe how the representation shown for the paramagnetic material would have to be changed if the material were placed in a magnetic field.
GIVE IT SOME THOUGHT Why doesn’t Ti5 + exist?
Magnetism The spin an electron possesses gives the electron a magnetic moment, a property that causes the electron to behave like a tiny (a) Paramagnetic; spins (b) Ferromagnetic; spins magnet. In a diamagnetic solid, defined as one in which all the random; spins do aligned; spins become electrons in the solid are paired, the spin-up and spin-down align if in magnetic random at high electrons cancel one another. • (Section 9.8) Diamagnetic field temperature substances are generally described as being nonmagnetic, but when a diamagnetic substance is placed in a magnetic field, the motions of the electrons cause the substance to be very weakly repelled by the magnet. In other words, these supposedly nonmagnetic substances do show some very faint magnetic character in the presence of a magnetic field. A substance in which the atoms or ions have one or more unpaired electrons is paramagnetic. •(Section 9.8) In a paramagnetic solid, the electrons on one atom or ion do not influence the unpaired electrons on neighboring atoms or ions. As a result, the magnetic moments on the atoms or ions are (d) Ferrimagnetic; unequal (c) Antiferromagnetic; spins opposed and spins opposed but do randomly oriented, as shown in 씰 FIGURE 23.5(a). When a cancel; spins become not cancel; spins become paramagnetic substance is placed in a magnetic field, however, random at high random at high the magnetic moments tend to align parallel to one another, temperature temperature producing a net attractive interaction with the magnet. Thus, 쑿 FIGURE 23.5 The relative orientation of electron spins in unlike a diamagnetic substance, which is weakly repulsed by a various types of magnetic substances. magnetic field, a paramagnetic substance is attracted to a magnetic field. When you think of a magnet, you probably picture a simple iron magnet. Iron exhibits ferromagnetism, a form of magnetism much stronger than paramagnetism. Ferromagnetism arises when the unpaired electrons of the atoms or ions in a solid are influenced by the orientations of the electrons in neighboring atoms or ions. The most stable (lowest-energy) arrangement is when the spins of electrons on adjacent atoms or ions are aligned in the same direction, as in Figure 23.5(b). When a ferromagnetic solid is placed in a magnetic field, the electrons tend to align strongly in a direction parallel to the magnetic field. The attraction to the magnetic field that results may be as much as one million times stronger than that for a paramagnetic substance. When a ferromagnet is removed from an external magnetic field, the interactions between the electrons cause the ferromagnetic substance to maintain a magnetic moment. We then refer to it as a permanent magnet (씰 FIGURE 23.6). The only ferromagnetic transition metals are Fe, Co, and Ni, but many alloys also exhibit ferromagnetism, which is in some cases stronger than the ferromagnetism of the pure metals. Particularly powerful ferromagnetism is found in compounds containing both transition metals and lanthanide metals. Two of the most important examples are SmCo5 and Nd2Fe14B. Two additional types of magnetism involving ordered arrangements of unpaired electrons are depicted in Figure 23.5. In materials that exhibit antiferromagnetism 쑿 FIGURE 23.6 A permanent magnet. Permanent magnets are made from [Figure 23.5(c)], the unpaired electrons on a given atom or ion align so that their ferromagnetic and ferrimagnetic materials. spins are oriented in the direction opposite the spin direction on neighboring atoms.
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Transition Metals and Coordination Chemistry
This means that the spin-up and spin-down electrons cancel each other. Examples of antiferromagnetic substances are chromium metal, FeMn alloys, and such transitionmetal oxides as Fe2O3, LaFeO3, and MnO. A substance that exhibits ferrimagnetism [Figure 23.5(d)] has both ferromagnetic and antiferromagnetic properties. Like an antiferromagnet, the unpaired electrons align so that the spins in adjacent atoms or ions point in opposite directions. However, unlike an antiferromagnet, the net magnetic moments of the spin-up electrons are not fully canceled by the spin-down electrons. This can happen because the magnetic centers have different numbers of unpaired electrons (NiMnO3), because the number of magnetic sites aligned in one direction is larger than the number aligned in the other direction (Y3Fe5O12), or because both these conditions apply (Fe3O4). Because the magnetic moments do not cancel, the properties of ferrimagnetic materials are similar to the properties of ferromagnetic materials. GIVE IT SOME THOUGHT How do you think spin-spin interactions of unpaired electrons on adjacent atoms in a substance are affected by the interatomic distance?
Ferromagnets, ferrimagnets, and antiferromagnets all become paramagnetic when heated above a critical temperature. This happens when the thermal energy is sufficient to overcome the forces determining the spin directions of the electrons. This temperature is called the Curie temperature, TC, for ferromagnets and ferrimagnets and the Néel temperature, TN, for antiferromagnets.
23.2 | TRANSITION-METAL COMPLEXES The transition metals occur in many interesting and important molecular forms. Species that are assemblies of a central transition-metal ion bonded to a group of surrounding molecules or ions, such as 3Ag(NH3)24 + and 3Fe(H2O)643 + , are called metal complexes, or merely complexes.* If the complex carries a net charge, it is generally called a complex ion. • (Section 17.5) Compounds that contain complexes are known as coordination compounds. The molecules or ions that bond to the metal ion in a complex are known as ligands (from the Latin word ligare, “to bind”). There are two NH3 ligands bonded to Ag + in the complex ion 3Ag(NH3)24 + , for instance, and six H2O ligands bonded to Fe3+ in 3Fe(H2O)643 + . Each ligand functions as a Lewis base and so donates a pair of electrons to form the ligand–metal bond. •(Section 16.11) Thus, every ligand has at least one unshared pair of valence electrons. Four of the most frequently encountered ligands, H O H
H
N
H
Cl
⫺
C
N⫺
H
illustrate that most ligands are either polar molecules or anions. In forming a complex, the ligands are said to coordinate to the metal. GIVE IT SOME THOUGHT Is the interaction between an ammonia ligand and a metal cation a Lewis acid–base interaction? If so, which species acts as the Lewis acid?
*Most of the coordination compounds we examine in this chapter contain transition-metal ions, although ions of other metals can also form complexes.
SECTION 23.2
TABLE 23.3 • Properties of Some Ammonia Complexes of Cobalt(III) Original Formulation
Color
Ions per Formula Unit
“Free” Cl ⴚ Ions per Formula Unit
Modern Formulation
CoCl3 · 6 NH3 CoCl3 · 5 NH3 CoCl3 · 4 NH3 CoCl3 · 4 NH3
Orange Purple Green Violet
4 3 2 2
3 2 1 1
[Co(NH3)6]Cl3 [Co(NH3)5Cl]Cl2 trans-[Co(NH3)4Cl2]Cl cis-[Co(NH3)4Cl2]Cl
The Development of Coordination Chemistry: Werner’s Theory Because compounds of the transition metals are beautifully colored, the chemistry of these elements fascinated chemists even before the periodic table was introduced. During the late 1700s through the 1800s, the many coordination compounds that were isolated and studied had properties that were puzzling in light of the bonding theories prevailing at the time. 쑿 TABLE 23.3, for example, lists a series of CoCl3-NH3 compounds that have strikingly different colors. Note that the third and fourth species have different colors even though the originally assigned formula was the same for both, CoCl 3 # 4 NH3. The modern formulations of the compounds in Table 23.3 are based on various lines of experimental evidence. For example, all four compounds are strong electrolytes • (Section 4.1) but yield different numbers of ions when dissolved in water. Dissolving CoCl 3 # 6 NH3 in water yields four ions per formula unit (3Co(NH3)643 + plus three Cl- ions), whereas CoCl 3 # 5 NH3 yields only three ions per formula unit (3Co(NH3)5Cl42 + and two Cl- ions). Furthermore, the reaction of the compounds with excess aqueous silver nitrate leads to the precipitation of different amounts of AgCl(s). When CoCl 3 # 6 NH3 is treated with excess AgNO3(aq), 3 mol of AgCl(s) are produced per mole of complex, which means all three Cl- ions in the complex can react to form AgCl(s). By contrast, when CoCl 3 # 5 NH3 is treated with excess AgNO3(aq), only 2 mol of AgCl(s) precipitate per mole of complex, telling us that one of the Cl- ions in the complex does not react. These results are summarized in Table 23.3. In 1893 the Swiss chemist Alfred Werner (1866–1919) proposed a theory that successfully explained the observations in Table 23.3. In a theory that became the basis for understanding coordination chemistry, Werner proposed that any metal ion exhibits both a primary valence and a secondary valence. The primary valence is the oxidation state of the metal, which is +3 for the complexes in Table 23.3. • (Section 4.4) The secondary valence is the number of atoms bonded to the metal ion, which is also called the coordination number. For these cobalt complexes, Werner deduced a coordination number of 6 with the ligands in an octahedral arrangement around the Co3 + ion. Werner’s theory provided a beautiful explanation for the results in Table 23.3. The NH3 molecules are ligands bonded to the Co3 + ion; if there are fewer than six NH3 molecules, the remaining ligands are Cl- ions. The central metal and the ligands bound to it constitute the coordination sphere of the complex. In writing the chemical formula for a coordination compound, Werner suggested using square brackets to signify the makeup of the coordination sphere in any given compound. He therefore proposed that CoCl 3 # 6 NH3 and CoCl 3 # 5 NH3 are better written as [Co(NH3)6]Cl3 and [Co(NH3)5Cl]Cl2, respectively. He further proposed that the chloride ions that are part of the coordination sphere are bound so tightly that they do not dissociate when the complex is dissolved in water. Thus, dissolving [Co(NH3)5Cl]Cl2 in water produces a 3Co(NH3)5Cl42 + ion and two Cl- ions. Werner’s ideas also explained why there are two forms of CoCl 3 # 4 NH3. Using Werner’s postulates, we write the formula as [Co(NH3)4Cl2]Cl. As shown in 씰 FIGURE 23.7, there are two ways to arrange the ligands in the 3Co(NH3)4Cl 24 + complex, called the cis and trans forms. In the cis form, the two chloride ligands occupy adjacent vertices
Transition-Metal Complexes
969
970
CHAPTER 23
Transition Metals and Coordination Chemistry
⫹
⫹
Two Cl on same side of metal ion 씰 FIGURE 23.7 Isomers of [Co(NH3)4Cl2]ⴙ. The cis isomer is violet, and the trans isomer is green.
cis isomer
Two Cl on opposite sides of metal ion
trans isomer
of the octahedral arrangement. In trans-3Co(NH3)4Cl 24 + the two chlorides are opposite each other. It is this difference in positions of the Cl ligands that leads to two compounds, one violet and one green. The insight Werner provided into the bonding in coordination compounds is even more remarkable when we realize that his theory predated Lewis’s ideas of covalent bonding by more than 20 years! Because of his tremendous contributions to coordination chemistry, Werner was awarded the 1913 Nobel Prize in Chemistry. SAMPLE EXERCISE 23.1
Identifying the Coordination Sphere of a Complex
Palladium(II) tends to form complexes with coordination number 4. A compound has the composition PdCl 2 # 3 NH3. (a) Write the formula for this compound that best shows the coordination structure. (b) When an aqueous solution of the compound is treated with excess AgNO3(aq), how many moles of AgCl(s) are formed per mole of PdCl 2 # 3 NH3? SOLUTION Analyze We are given the coordination number of Pd(II) and a chemical formula indicating that the complex contains NH3 and Cl- . We are asked to determine (a) which ligands are attached to Pd(II) in the compound and (b) how the compound behaves toward AgNO3 in aqueous solution. Plan (a) Because of their charge, the Cl- ions can be either in the coordination sphere, where they are bonded directly to the metal, or outside the coordination sphere, where they are bonded ionically to the complex. The electrically neutral NH3 ligands must be in the coordination sphere, if we assume four ligands bonded to the Pd(II) ion. (b) Any chlorides in the coordination sphere do not precipitate as AgCl. Solve (a) By analogy to the ammonia complexes of cobalt(III) shown in Figure 23.7, we predict that the three NH3 are ligands attached to the Pd(II) ion. The fourth ligand around Pd(II) is one chloride ion. The second chloride ion is not a ligand; it serves only as a counterion (a noncoordinating ion that balances charge) in the compound. We conclude that the formula showing the structure best is [Pd(NH3)3Cl]Cl. (b) Because only the non-ligand Cl- can react, we expect to produce 1 mol of AgCl(s) per mole of complex. The balanced equation is 3Pd(NH3)3Cl4Cl(aq) + AgNO3(aq) ¡ 3Pd(NH3)3Cl4NO3(aq) + AgCl(s)
SECTION 23.2
Transition-Metal Complexes
971
This is a metathesis reaction •(Section 4.2) in which one of the cations is the 3Pd(NH3)3Cl4 + complex ion. PRACTICE EXERCISE Predict the number of ions produced per formula unit in an aqueous solution of CoCl 2 # 6 H2O. Answer: three: 3Co(H2O)642 + and two Cl-
The Metal–Ligand Bond The bond between a ligand and a metal ion is a Lewis acid–base interaction. Because the ligands have available pairs of electrons, they can function as Lewis bases (electron-pair donors). Metal ions (particularly transition-metal ions) have empty valence orbitals, so they can act as Lewis acids (electron-pair acceptors). We can picture the bond between the metal ion and ligand as the result of their sharing a pair of electrons initially on the ligand: H
H Ag⫹(aq) ⫹ 2 N
H(aq)
H
N Ag N H
H
H
⫹
H (aq)
[23.1]
H
The formation of metal–ligand bonds can markedly alter the properties we observe for the metal ion. A metal complex is a distinct chemical species that has physical and chemical properties different from those of the metal ion and ligands from which it is formed. As one example, 쑼 FIGURE 23.8 shows the color change that occurs when aqueous solutions of NCS- (colorless) and Fe3+ (yellow) are mixed, forming 3Fe(H2O)5NCS42 + . Complex formation can also significantly change other properties of metal ions, such as their ease of oxidation or reduction. Silver ion, for example, is readily reduced in water, Ag + (aq) + e - ¡ Ag(s)
E° = +0.799 V
[23.2]
but the 3Ag(CN)24- ion is not so easily reduced because complexation by CN - ions stabilizes silver in the +1 oxidation state: 3Ag(CN)24-(aq) + e - ¡ Ag(s) + 2 CN -(aq)
E ° = -0.31 V
[23.3]
Hydrated metal ions are complexes in which the ligand is water. Thus, Fe 3 + (aq) consists largely of 3Fe(H2O)643 + . • (Section 16.11) It is important to realize that NH4NCS(aq) solution
[Fe(H2O)6]3⫹(aq) solution
Red [Fe(H2O)5NCS]2⫹ forms
씱 FIGURE 23.8 Reaction of Fe3ⴙ(aq) and NCSⴚ(aq).
972
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Transition Metals and Coordination Chemistry
ligands can undergo reaction. For example, we saw in Figure 16.16 that a water molecule in 3Fe(H2O)643 + (aq) can be deprotonated to yield 3Fe(H2O)5OH42 + (aq) and H + (aq). The iron ion retains its oxidation state; the coordinated hydroxide ligand, with a 1charge, reduces the complex charge to 2+ . Ligands can also be displaced from the coordination sphere by other ligands, if the incoming ligands bind more strongly to the metal ion than the original ones. For example, ligands such as NH3, NCS-, and CN- can replace H2O in the coordination sphere of metal ions. GIVE IT SOME THOUGHT Write a balanced chemical equation for the reaction that causes the color change in Figure 23.8.
Charges, Coordination Numbers, and Geometries The charge of a complex is the sum of the charges on the metal and on the ligands. In [Cu(NH3)4]SO4 we can deduce the charge on the complex ion because we know that the charge of the sulfate ion is 2- . Because the compound is electrically neutral, the complex ion must have a 2+ charge, 3Cu(NH3)442 + . We can then use the charge of the complex ion to deduce the oxidation number of copper. Because the NH3 ligands are uncharged molecules, the oxidation number of copper must be +2: ⫹2 ⫹ 4(0) ⫽ ⫹2 [Cu(NH3)4]2⫹ SAMPLE EXERCISE 23.2
Determining the Oxidation Number of a Metal in a Complex
What is the oxidation number of the metal in [Rh(NH3)5Cl](NO3)2? SOLUTION Analyze We are given the chemical formula of a coordination compound and asked to determine the oxidation number of its metal atom. Plan To determine the oxidation number of Rh, we need to figure out what charges are contributed by the other groups. The overall charge is zero, so the oxidation number of the metal must balance the charge due to the rest of the compound. Solve The NO3 group is the nitrate anion, which has a 1- charge. The NH3 ligands carry zero charge, and the Cl is a coordinated chloride ion, which has a 1- charge. The sum of all the charges must be zero: x ⫹ 5(0) ⫹ (⫺1) ⫹ 2(⫺1) ⫽ 0 [Rh(NH3)5Cl](NO3)2
The oxidation number of rhodium, x, must therefore be +3. PRACTICE EXERCISE What is the charge of the complex formed by a platinum(II) metal ion surrounded by two ammonia molecules and two bromide ions? Answer: zero
SAMPLE EXERCISE 23.3
Determining the Formula of a Complex Ion
A complex ion contains a chromium(III) bound to four water molecules and to two chloride ions. What is the formula and charge of this ion? SOLUTION Analyze We are given a metal, its oxidation number, and the number of ligands of each kind in a complex ion containing the metal, and we are asked to write the chemical formula and charge of the ion.
SECTION 23.2
Transition-Metal Complexes
973
Plan We write the metal first, then the ligands. We can use the charges of the metal ion and ligands to determine the charge of the complex ion. The oxidation state of the metal is +3, water has no charge, and chloride has a 1 - charge. Solve ⫹3 ⫹ 4(0) ⫹ 2(⫺1) ⫽ ⫹1 Cr(H2O)4Cl2
The charge on the ion is 1 + , 3Cr(H2O)4Cl 24 + . PRACTICE EXERCISE Write the formula for the complex described in the Practice Exercise accompanying Sample Exercise 23.2. Answer: [Pt(NH3)2Br2]
Recall that the number of atoms directly bonded to the metal atom in a complex is the coordination number of the complex. Thus, the silver ion in 3Ag(NH3)24 + has a coordination number of 2, and the cobalt ion has a coordination number of 6 in all four complexes in Table 23.3. Some metal ions have only one observed coordination number. The coordination number of chromium(III) and cobalt(III) is invariably 6, for example, and that of platinum(II) is always 4. For most metals, however, the coordination number is different for different ligands. In these complexes, the most common coordination numbers are 4 and 6. The coordination number of a metal ion is often influenced by the relative sizes of the metal ion and the ligands. As the ligand gets larger, fewer of them can coordinate to the metal ion. Thus, iron(III) is able to coordinate to six fluorides in 3FeF643- but to only four chlorides in 3FeCl 44- . Ligands that transfer substantial negative charge to the metal also GO FIGURE produce reduced coordination numbers. For example, six ammonia What is the size of the NH3-Zn-NH3 bond angle? molecules can coordinate to nickel(II), forming 3Ni(NH3)642 +, but only Of the NH3-Pt-NH3 bond angle? 2four cyanide ions can coordinate to this ion, forming 3Ni(CN)44 . Complexes in which the coordination number is 4 have two com2⫹ 2⫹ NH3 H3N mon geometries—tetrahedral and square planar (씰 FIGURE 23.9). The tetrahedral geometry is the more common of the two and is espeNH3 H3N Pt Zn NH3 H3N cially common among nontransition metals. The square-planar geometry is characteristic of transition-metal ions with eight d elecNH3 NH3 trons in the valence shell, such as platinum(II) and gold(III). The vast majority of complexes in which the coordination number tetrahedral square planar is 6 have an octahedral geometry. The octahedron, which has eight faces and six vertices, is often represented as a square with ligands 쑿 FIGURE 23.9 In complexes having coordination above and below the plane, as in 쑼 FIGURE 23.10. Recall, however, number 4, the molecular geometry can be tetrahedral or square planar. that all six vertices on an octahedron are geometrically equivalent. 3⫹
NH3
H3N H3N
3⫹
NH3
H3N H3N
Co
H3N
Co
NH3
NH3
NH3 NH3
NH3
씱 FIGURE 23.10 In complexes having coordination number 6, the molecular geometry is almost always octahedral. Two ways to draw octahedral geometry are shown.
974
CHAPTER 23
Transition Metals and Coordination Chemistry • (Section 9.2) The octahedron can also be thought of as two square pyramids that
share the same square base. GIVE IT SOME THOUGHT What are the geometries most commonly associated with a. coordination number 4, b. coordination number 6?
|
23.3 COMMON LIGANDS IN COORDINATION CHEMISTRY The ligand atom that binds to the central metal ion in a coordination complex is called the donor atom of the ligand. Ligands having only one donor atom are called monodentate ligands (from the Latin, meaning “one-toothed”). These ligands are able to occupy only one site in a coordination sphere. Ligands having two donor atoms are bidentate ligands (“two-toothed”), and those having three or more donor atoms are polydentate ligands (“many-toothed”). In both bidentate and polydentate species, the multiple donor atoms can simultaneously bond to the metal ion, thereby occupying two or more sites in a coordination sphere. 쑼 TABLE 23.4 gives examples of all three types of ligands. Because they appear to grasp the metal between two or more donor atoms, bidentate and polydentate ligands are also known as chelating agents (pronounced “KEE-lay-ting”; from the Greek chele, “claw”).
TABLE 23.4 • Some Common Ligands Ligand Type
Examples
Monodentate
H2O
F
Water
NH3 Ammonia
Cl
⫺ ⫺
Fluoride ion
[C
N ]⫺
Chloride ion
[S
C or
Cyanide ion
N ]⫺ Thiocyanate ion
[O
H ]⫺
[O
N
O H2N
CH2 N
Ethylenediamine (en)
N
N
O
H2N
CH2
CH2
O
CH2 NH2
NH
P
O O
O
Diethylenetriamine
P
O
O
P O
Triphosphate ion
C C
4⫺
O CH2 N
O
O
CH2
CH2
CH2
CH2
C
O
CH2
C
O
N
O
O 4⫺
Ethylenediaminetetraacetate ion (EDTA )
2⫺
O
2⫺
O
C
Oxalate ion
5⫺
O
O O
O
N
Ortho-phenanthroline (o-phen)
Bipyridine (bipy or bpy)
Polydentate
H2C
O C
NH2
O ]⫺ Nitrite ion
or
Bidentate
H2C
Hydroxide ion
O
C
O
Carbonate ion
SECTION 23.3
Common Ligands in Coordination Chemistry
975
3⫹ 3⫹
H2 C H2C H2C
H2 H2N N Co N H2 H N 2
CH2
NH2
⫽
NH2 C H2
CH2 씱 FIGURE 23.11 The [Co(en)3]3ⴙ ion. The ligand is ethylenediamine.
[Co(en)3]3⫹
One common chelating agent is the bidentate ligand ethylenediamine, en: CH2 H2N
CH2 NH2
in which each donor nitrogen atom has one nonbonding electron pair. These donor atoms are sufficiently far apart to allow both of them to bond to the metal ion in adjacent positions. The 3Co(en)343 + complex ion, which contains three ethylenediamine ligands in the octahedral coordination sphere of cobalt(III), is shown in 쑿 FIGURE 23.11. Notice that in the image on the right the en is written in a shorthand notation as two nitrogen atoms connected by an arc. The ethylenediaminetetraacetate ion, 3EDTA44-, is an important polydentate ligand that has six donor atoms. It can wrap around a metal ion using all six donor atoms, as shown in 씰 FIGURE 23.12, although it sometimes binds to a metal using only five of its donor atoms. In general, the complexes formed by chelating ligands (that is, bidentate and polydentate ligands) are more stable than the complexes formed by related monodentate ligands. The equilibrium formation constants for 3Ni(NH3)642 + and 3Ni(en)342 + illustrate this observation: 3Ni(H2O)642 + (aq) + 6 NH3(aq) Δ 3Ni(NH3)642 + (aq) + 6 H2O(l) Kf = 1.2 * 109 2+
O
O ⫺
O
CCH2
CH2C
O⫺
CH2C
O⫺
NCH2CH2N ⫺
O
CCH2 O
[EDTA]4⫺
O ⫺
[23.4]
2+
3Ni(H2O)64 (aq) + 3 en(aq) Δ 3Ni(en)34 (aq) + 6 H2O(l) Kf = 6.8 * 1017
[23.5]
Although the donor atom is nitrogen in both instances, 3Ni(en)342 + has a formation constant that is more than 108 times larger than that of 3Ni(NH3)642 + . This trend of generally larger formation constants for bidentate and polydentate ligands, known as the chelate effect, is examined in the “A Closer Look” essay on page 977. Chelating agents are often used to prevent one or more of the customary reactions of a metal ion without removing the ion from solution. For example, a metal ion that interferes with a chemical analysis can often be complexed and its interference thereby removed. In a sense, the chelating agent hides the metal ion. For this reason, scientists sometimes refer to these ligands as sequestering agents. Phosphate ligands, such as sodium tripolyphosphate, Na5[OPO2OPO2OPO3], are used to sequester Ca2 + and Mg 2 + ions in hard water so that these ions cannot interfere with the action of soap or detergents. Chelating agents are used in many prepared foods, such as salad dressings and frozen desserts, to complex trace metal ions that catalyze decomposition reactions. Chelating agents are used in medicine to remove toxic heavy metal ions that have been ingested, such as Hg 2 + , Pb 2 + , and Cd2 + . One method of treating lead poisoning, for example, is to administer Na23Ca(EDTA)4. The EDTA chelates the lead, allowing it to be removed from the body via urine.
쑿 FIGURE 23.12 The EDTA4ⴚ ligand (top) and the complex ion [Co(EDTA)]ⴚ (bottom). The ligand is the polydentate ethylenediaminetetraacetate ion, which has six donor atoms, two N and four O.
976
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Transition Metals and Coordination Chemistry
GIVE IT SOME THOUGHT
GO FIGURE
What is the coordination number of the metal ion in heme b? In chlorophyll a? N
NH N
Cobalt(III) has a coordination number of 6 in all its complexes. Is the carbonate ion a monodentate or bidentate ligand in the 3Co(NH3)4CO 34 + ion?
Metals and Chelates in Living Systems
HN
Porphine
N
N Fe2⫹ N
O
N
OH
OH
O
Heme b CH2
CH3
H3C N
N Mg N
N CH3
H3C CH3 H 3C
CH3
CH3 2
O O
CH3
2⫹
O
O OCH3
Chlorophyll a 쑿 FIGURE 23.13 Porphine and two porphyrins, heme b and chlorophyll a. Fe(II) and Mg(II) ions replace the two blue H atoms in porphine and bond with all four nitrogens in heme b and chlorophyll a, respectively.
Ten of the 29 elements known to be necessary for human life are transition metals. • (Section 2.7, “Chemistry and Life: Elements Required by Living Organisms”) These ten elements—V, Cr, Mn, Fe, Co, Ni, Cu, Zn, Mo, and Cd—form complexes with a variety of groups present in biological systems. Although our bodies require only small quantities of metals, deficiencies can lead to serious illness. A deficiency of manganese, for example, can lead to convulsive disorders. Some epilepsy patients have been helped by the addition of manganese to their diets. Among the most important chelating agents in nature are those derived from the porphine molecule (씱 FIGURE 23.13). This molecule can coordinate to a metal via its four nitrogen donor atoms. Once porphine bonds to a metal ion, the two H atoms on the nitrogens are displaced to form complexes called porphyrins. Two important porphyrins are hemes, in which the metal ion is Fe(II), and chlorophylls, with a Mg(II) central ion. 쑼 FIGURE 23.14 shows a schematic structure of myoglobin, a protein that contains one heme group. Myoglobin is a globular protein, one that folds into a compact, roughly spherical shape. Myoglobin is found in the cells of skeletal muscle, particularly in seals, whales, and porpoises. It stores oxygen in cells, one molecule of O2 per myoglobin, until it is needed for metabolic activities. Hemoglobin, the protein that transports oxygen in human blood, is made up of four heme-containing subunits, each of which is very similar to myoglobin. One hemoglobin can bind up to four O2 molecules. In both myoglobin and hemoglobin, the iron is coordinated to the four nitrogen atoms of a porphyrin and to a nitrogen atom from the protein chain (씰 FIGURE 23.15). In hemoglobin, the sixth position around the iron is occupied either by O2 (in oxyhemoglobin, the Heme
쑿 FIGURE 23.14 Myoglobin. This ribbon diagram does not show most of the atoms.
SECTION 23.3
Common Ligands in Coordination Chemistry
bright red form) or by water (in deoxyhemoglobin, the purplish red form). (The oxy form is the one shown in Figure 23.15.) Carbon monoxide is poisonous because the equilibrium binding constant of human hemoglobin for CO is about 210 times greater than that for O2. As a result, a relatively small quantity of CO can inactivate a substantial fraction of the hemoglobin in the blood. For example, a person breathing air that contains only 0.1% CO takes in enough CO after a few hours to convert up to 60% of the hemoglobin (Hb) into COHb, thereby reducing the blood’s normal oxygen-carrying capacity by 60%. Under normal conditions, a nonsmoker breathing unpolluted air has about 0.3 to 0.5% COHb in her or his blood. This amount arises mainly from the production of small quantities of CO in the course of normal body chemistry and from the small amount of CO present in clean air. Exposure to higher concentrations of CO causes the COHb level to increase, which in turn leaves fewer Hb sites to which O2 can bind. If the level of COHb becomes too high, oxygen transport is effectively shut down and death occurs. Because CO is colorless and odorless, CO poisoning occurs with very little warning. Improperly ventilated combustion devices, such as kerosene lanterns and stoves, thus pose a potential health hazard.
977
GO FIGURE
Where would CO bind in this molecule? O
O Heme
N HN Protein (globin) 쑿 FIGURE 23.15 Coordination sphere of the hemes in oxymyoglobin and oxyhemoglobin.
A CLOSER LOOK ENTROPY AND THE CHELATE EFFECT We learned in Section 19.5 that chemical processes are favored by positive entropy changes and by negative enthalpy changes. The special stability associated with the formation of chelates, called the chelate effect, can be explained by comparing the entropy changes that occur with monodentate ligands with the entropy changes that occur with polydentate ligands. We begin with the reaction in which two H2O ligands of the square-planar Cu(II) complex 3Cu(H2O)442 + are replaced by monodentate NH3 ligands at 27 °C: 3Cu(H2O)442 + (aq) + 2 NH3(aq) Δ 3Cu(H2O)2(NH3)242 + (aq) + 2 H2O(l) ¢H° = -46 kJ;
¢S° = -8.4 J>K;
¢G° = -43 kJ
The thermodynamic data tell us about the relative abilities of H2O and NH3 to serve as ligands in this reaction. In general, NH3 binds more tightly to metal ions than does H2O, so this substitution reaction is exothermic (¢H 6 0). The stronger bonding of the NH3 ligands also causes the 3Cu(H2O)2(NH3)242 + ion to be more rigid, which is probably the reason ¢S° is slightly negative. We can use Equation 19.20, ¢G° = -RT ln K, to calculate the equilibrium constant of the reaction at 27 °C. The result, K = 3.1 * 107, tells us that the equilibrium lies far to the right, favoring replacement of H2O by NH3. For this equilibrium, therefore, the enthalpy change, ¢H° = -46 kJ, is large enough and negative enough to overcome the entropy change, ¢S° = -8.4 J>K. Now let’s use a single bidentate ethylenediamine ligand in our substitution reaction: 3Cu(H2O)442 + (aq) + en(aq) Δ 3Cu(H2O)2(en)42 + (aq) + 2 H2O(l) ¢H° = -54 kJ; ¢S° = +23 J>K; ¢G° = -61 kJ The en ligand binds slightly more strongly to the Cu2 + ion than two NH3 ligands, so the enthalpy change here (-54 kJ) is slightly more
negative than for 3Cu(H2O)2(NH3)242 + (-46 kJ). There is a big difference in the entropy change, however: ¢S° is -8.4 J> K for the NH3 reaction but +23 J> K for the en reaction. We can explain the positive ¢S° value by using concepts discussed in Section 19.3. Because a single en ligand occupies two coordination sites, two molecules of H2O are released when one en ligand bonds. Thus, there are three product molecules in the reaction but only two reactant molecules. The greater number of product molecules leads to the positive entropy change for the equilibrium. The slightly more negative value of ¢H° for the en reaction ( -54 kJ versus -46 kJ) coupled with the positive entropy change leads to a much more negative value of ¢G° (-61 kJ for en, -43 for NH3) and a correspondingly larger equilibrium constant: K = 4.2 * 1010. We can combine our two equations using Hess’s law •(Section 5.6) to calculate the enthalpy, entropy, and free-energy changes that occur for en to replace ammonia as ligands on Cu(II): 3Cu(H2O)2(NH3)242 + (aq) + en(aq) Δ 3Cu(H2O)2(en)42 + (aq) + 2 NH3(aq) ¢H° = (-54 kJ) - (-46 kJ) = -8 kJ ¢S° = (+23 J>K) - (-8.4 J>K) = +31 J>K ¢G° = (- 61 kJ) - (-43 kJ) = -18 kJ Notice that at 27 °C, the entropic contribution (-T¢S°) to the free-energy change, ¢G° = ¢H° - T¢S° (Equation 19.12), is negative and greater in magnitude than the enthalpic contribution (¢H°). The equilibrium constant for the NH3–en reaction, 1.4 * 103, shows that the replacement of NH3 by en is thermodynamically favorable. The chelate effect is important in biochemistry and molecular biology. The additional thermodynamic stabilization provided by entropy effects helps stabilize biological metal–chelate complexes, such as porphyrins, and can allow changes in the oxidation state of the metal ion while retaining the structural integrity of the complex. RELATED EXERCISES: 23.31, 23.32, 23.33, 23.98
978
CHAPTER 23
Transition Metals and Coordination Chemistry
GO FIGURE
Which peak in this curve corresponds to the lowest-energy transition by an electron in a chlorophyll molecule?
The chlorophylls, which are porphyrins that contain Mg(II) (Figure 23.13), are the key components in the conversion of solar energy into forms that can be used by living organisms. This process, called photosynthesis, occurs in the leaves of green plants: 6 CO2(g) + 6 H2O(l) ¡ C6H12O6(aq) + 6 O2(g)
[23.6]
Light absorption
The formation of 1 mol of glucose, C6H12O6, requires the absorption of 48 mol of photons from sunlight or other sources of light. Chlorophyll-containing pigments in the leaves of plants absorb the photons. Figure 23.13 shows that the chlorophyll molecule has a series of alternating, or conjugated, double bonds in the ring surrounding the metal ion. This system of conjugated double bonds makes it possible for chlorophyll to absorb light strongly in the visible region of the spectrum. As 씱 FIGURE 23.16 shows, chlorophyll is green because it absorbs red light (maximum absorption at 655 nm) and blue light (maximum absorption at 430 nm) and transmits green light. Photosynthesis is nature’s solar-energy–conversion machine, and thus all living systems on Earth depend on photosynthesis for continued existence. 400
500 600 Wavelength (nm)
700
쑿 FIGURE 23.16 The absorption of sunlight by chlorophyll.
GIVE IT SOME THOUGHT What property of the porphine ligand makes it possible for chlorophyll to play a role in plant photosynthesis?
CHEMISTRY AND LIFE THE BATTLE FOR IRON IN LIVING SYSTEMS Because living systems have difficulty assimilating enough iron to satisfy their nutritional needs, irondeficiency anemia is a common problem in humans. Chlorosis, an iron deficiency in plants that makes leaves turn yellow, is also commonplace. Living systems have difficulty assimilating iron because most iron compounds found in nature are not very soluble in water. Microorganisms have adapted to this problem by secreting an iron-binding compound, called a siderophore, that forms an extremely stable water-soluble complex with iron(III). One such complex is ferrichrome (씰 FIGURE 23.17). The iron-binding strength of a siderophore is so great that it can extract iron from glass cooking pots and the iron in iron oxides. When ferrichrome enters a living cell, the iron it carries is removed through an enzyme-catalyzed reaction that reduces the strongly bonding iron(III) to iron(II), which is only weakly complexed by the siderophore (씰 FIGURE 23.18). Microorganisms thus acquire iron by excreting a siderophore into their immediate environment and then taking the resulting iron complex into the cell. In humans, iron is assimilated from food in the intestine. A protein called transferrin binds iron and transports it across the intestinal wall to distribute it to other tissues in the body. The normal adult body contains about 4 g of iron. At any one time, about 3 g of this iron is in the blood, mostly in the form of hemoglobin. Most of the remainder is carried by transferrin. A bacterium that infects the blood requires a source of iron if it is to grow and reproduce. The bacterium excretes a siderophore into the blood to compete with transferrin for iron. The equilibrium con-
H
H N
O
H
C
CH2 CH2
C H
O
C
N
CH2
C
O
C
N
O
CH2 CH2
O
N
Fe O
C
O
C C
H CH3
C
CH3
C N
O N
H N
CH2 CH2
C O
C H
CH2 C
O
H
H
C
H
H
H
CH2 C
N
H CH3
3⫹
O
H
O
N H
쑿 FIGURE 23.17 Ferrichrome.
stants for forming the iron complex are about the same for transferrin and siderophores. The more iron available to the bacterium, the more rapidly it can reproduce and thus the more harm it can do. Several years ago, New Zealand clinics regularly gave iron supplements to infants soon after birth. However, the incidence of certain bacterial infections was eight times higher in treated than in
SECTION 23.4
Nomenclature and Isomerism in Coordination Chemistry
Cell wall
untreated infants. Presumably, the presence of more iron in the blood than absolutely necessary makes it easier for bacteria to obtain the iron needed for growth and reproduction. In the United States it is common medical practice to supplement infant formula with iron sometime during the first year of life. However, iron supplements are not necessary for infants who breastfeed because breast milk contains two specialized proteins, lactoferrin and transferrin, which provide sufficient iron while denying its availability to bacteria. Even for infants fed with infant formulas, supplementing with iron during the first several months of life may be ill-advised. For bacteria to continue to multiply in the blood, they must synthesize new supplies of siderophores. Synthesis of siderophores in bacteria slows, however, as the temperature is increased above the normal body temperature of 37 °C and stops completely at 40 °C. This suggests that fever in the presence of an invading microbe is a mechanism used by the body to deprive bacteria of iron.
Siderophore Fe3⫹
Fe2⫹
Fe3⫹
e⫺
Ferrichrome Fe3⫹
쑿 FIGURE 23.18 The iron-transport system of a bacterial cell.
RELATED EXERCISE: 23.74
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23.4 NOMENCLATURE AND ISOMERISM IN COORDINATION CHEMISTRY When complexes were first discovered, they were named after the chemist who originally prepared them. A few of these names persist, as, for example, with the dark red substance NH4[Cr(NH3)2(NCS)4], which is still known as Reinecke’s salt. Once the structures of complexes were more fully understood, it became possible to name them in a more systematic manner. Let’s use two substances to illustrate how coordination compounds are named: [Co(NH3)5Cl]Cl2
Cation
Anion
Pentaamminechlorocobalt(III)
Chloride
5 NH3 ligands
Na2[MoOCl4]
Cl⫺ ligand
Cobalt in ⫹3 oxidation state
Cation
Anion
Sodium
Tetrachlorooxomolybdate(IV)
4 Cl⫺ ligands
979
Oxide, O2⫺ ligand
Molybdenum in ⫹4 oxidation state
1. In naming complexes that are salts, the name of the cation is given before the name of the anion. Thus, in [Co(NH3)5Cl]Cl2 we name the 3Co(NH3)5Cl42 + cation and then the Cl- . 2. In naming complex ions or molecules, the ligands are named before the metal. Ligands are listed in alphabetical order, regardless of their charges. Prefixes that give the number of ligands are not considered part of the ligand name in determining alphabetical order. Thus, the 3Co(NH3)5Cl42 + ion is pentaamminechlorocobalt(III). (Be careful to note, however, that the metal is written first in the chemical formula.)
980
CHAPTER 23
Transition Metals and Coordination Chemistry TABLE 23.5 • Some Common Ligands and Their Names Ligand
Name in Complexes Azido Bromo Chloro Cyano Fluoro Hydroxo
Ligand
Azide, N3Bromide, Br Chloride, CI Cyanide, CN Fluoride, F Hydroxide, OH -
Oxalate, C2O42Oxide, O2 Ammonia, NH3 Carbon monoxide, CO Ethylenediamine, en Pyridine, C5H5N
Name in Complexes Oxalato Oxo Ammine Carbonyl Ethylenediamine Pyridine
Carbonate, CO32-
Carbonato
Water, H2O
Aqua
3. The names of anionic ligands end in the letter o, but electrically neutral ligands ordinarily bear the name of the molecules (쑿 TABLE 23.5). Special names are used for H2O (aqua), NH3 (ammine), and CO (carbonyl). For example, 3Fe(CN)2(NH3)2(H2O)24 + is the diamminediaquadicyanoiron(III) ion. 4. Greek prefixes (di-, tri-, tetra-, penta-, hexa-) are used to indicate the number of each kind of ligand when more than one is present. If the ligand contains a Greek prefix (for example, ethylenediamine) or is polydentate, the alternate prefixes bis-, tris-, tetrakis-, pentakis-, and hexakis- are used and the ligand name is placed in parentheses. For example, the name for [Co(en)3]Br3 is tris(ethylenediamine)cobalt(III) bromide. 5. If the complex is an anion, its name ends in -ate. The compound K4[Fe(CN)6] is potassium hexacyanoferrate(II), for example, and the ion 3CoCl 442- is tetrachlorocobaltate(II) ion. 6. The oxidation number of the metal is given in parentheses in Roman numerals following the name of the metal. Three examples for applying these rules are [Ni(NH3)6]Br2 [Co(en)2(H2O)(CN)]Cl2 Na2[MoOCl4] SAMPLE EXERCISE 23.4
Hexaamminenickel(II) bromide Aquacyanobis(ethylenediamine)cobalt(III) chloride Sodium tetrachlorooxomolybdate(IV)
Naming Coordination Compounds
Name the compounds (a) [Cr(H2O)4Cl2]Cl, (b) K4[Ni(CN)4]. SOLUTION Analyze We are given the chemical formulas for two coordination compounds and assigned the task of naming them. Plan To name the complexes, we need to determine the ligands in the complexes, the names of the ligands, and the oxidation state of the metal ion. We then put the information together following the rules listed in the text. Solve (a) The ligands are four water molecules—tetraaqua—and two chlo- ⫹3 ⫹ 4(0) ⫹ 2(⫺1) ⫹ (⫺1) ⫽ 0 ride ions—dichloro. By assigning all the oxidation numbers we know for this molecule, we see that the oxidation number of Cr is +3: [Cr(H2O)4Cl2]Cl Thus, we have chromium(III). Finally, the anion is chloride. The name of the compound is tetraaquadichlorochromium(III) chloride (b) The complex has four cyanide ion ligands, CN- , which means tetracyano, and the oxidation state of the nickel is zero:
4(⫹1) ⫹ 0 ⫹ 4(⫺1) ⫽ 0 K4[Ni(CN)4]
Because the complex is an anion, the metal is indicated as nickelate(0). Putting these parts together and naming the cation first, we have
potassium tetracyanonickelate(0)
PRACTICE EXERCISE Name the compounds (a) [Mo(NH3)3Br3]NO3, (b) (NH4)2[CuBr4]. (c) Write the formula for sodium diaquabis(oxalato)ruthenate(III). Answers: (a) triamminetribromomolybdenum(IV) nitrate, (b) ammonium tetrabromocuprate(II) (c) Na[Ru(H2O)2(C2O4)2]
SECTION 23.4
Nomenclature and Isomerism in Coordination Chemistry
981
Isomers (same formula, different properties)
Structural isomers (different bonds)
Coordinationsphere isomers
Stereoisomers (same bonds, different arrangements)
Linkage isomers
Geometric isomers
Optical isomers
씱 FIGURE 23.19 Forms of isomerism in coordination compounds.
Isomerism When two or more compounds have the same composition but a different arrangement of atoms, we call them isomers. • (Section 2.9) Here we consider two main kinds of isomers in coordination compounds: structural isomers (which have different bonds) and stereoisomers (which have the same bonds but different ways in which the ligands occupy the space around the metal center). Each of these classes also has subclasses, as shown in 쑿 FIGURE 23.19.
Structural Isomerism Many types of structural isomerism are known in coordination chemistry, including the two named in Figure 23.19: linkage isomerism and coordination-sphere isomerism. Linkage isomerism is a relatively rare but interesting type that arises when a particular ligand is capable of coordinating to a metal in two ways. The nitrite ion, NO2 -, for example, can coordinate to a metal ion through either its nitrogen or one of its oxygens (씰 FIGURE 23.20). When it coordinates through the nitrogen atom, the NO2 - ligand is called nitro; when it coordinates through the oxygen atom, it is called nitrito and is generally written ONO- . The isomers shown in Figure 23.20 have different properties. The nitro isomer is yellow, for example, whereas the nitrito isomer is red. Another ligand capable of coordinating through either of two donor atoms is thiocyanate, SCN- , whose potential donor atoms are N and S. Coordination-sphere isomers are isomers that differ in which species in the complex are ligands and which are outside the coordination sphere in the solid lattice. For example, three isomers have the formula CrCl3(H2O)6. When the ligands are six H2O and the chloride ions are in the crystal lattice (as counterions), we have the violet compound [Cr(H2O)6]Cl3. When the ligands are five H2O and one Cl-, with the sixth H2O and the two Cl- out in the lattice, we have the green compound 3Cr(H2O)5Cl4Cl 2 # H2O. The third isomer, 3Cr(H2O)4Cl 24Cl # 2 H2O, is also a green compound. In the two green compounds, either one or two water molecules have been displaced from the coordination sphere by chloride ions and occupy a site in the crystal lattice. GIVE IT SOME THOUGHT Can the ammonia ligand engage in linkage isomerism? Explain.
2⫹
Nitro isomer Bonding via ligand N atom 쑿 FIGURE 23.20 Linkage isomerism.
2⫹
Nitrito isomer Bonding via ligand O atom
982
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Transition Metals and Coordination Chemistry
GO FIGURE
Which of these isomers has a nonzero dipole moment? ⫽N
씰 FIGURE 23.21 Geometric isomerism.
⫽ Cl
cis Cl ligands adjacent to each other NH3 ligands adjacent to each other
⫽H
⫽ Pt
trans Cl ligands on opposite sides of central atom NH3 ligands on opposite sides of central atom
Stereoisomerism Stereoisomers have the same chemical bonds but different spatial arrangements. In the square-planar complex [Pt(NH3)2Cl2], for example, the chloro ligands can be either adjacent to or opposite each other (쑿 FIGURE 23.21). (We saw an earlier example of this type of isomerism in the cobalt complex of Figure 23.7, and we’ll return to that complex in a moment.) This form of stereoisomerism, in which the arrangement of the atoms is different but the same bonds are present, is called geometric isomerism. The isomer on the left in Figure 23.21, with like ligands in adjacent positions, is the cis isomer, and the isomer on the right, with like ligands across from one another, is the trans isomer. Geometric isomers generally have different physical properties and may also have markedly different chemical reactivities. For example, cis-[Pt(NH3)2Cl2], also called cisplatin, is effective in the treatment of testicular, ovarian, and certain other cancers, whereas the trans isomer is ineffective. This is because cisplatin forms a chelate with two nitrogens of DNA, displacing the chloride ligands. The chloride ligands of the trans isomer are too far apart to form the N-Pt-N chelate with DNA nitrogens. Geometric isomerism is also possible in octahedral complexes when two or more different ligands are present, as in the cis and trans tetraamminedichlorocobalt(III) ion in Figure 23.7. Because all the corners of a tetrahedron are adjacent to one another, cistrans isomerism is not observed in tetrahedral complexes. SAMPLE EXERCISE 23.5
Determining the Number of Geometric Isomers
The Lewis structure ≠C ‚ O≠ indicates that the CO molecule has two lone pairs of electrons. When CO binds to a transition-metal atom, it nearly always does so by using the C lone pair. How many geometric isomers are there for tetracarbonyldichloroiron(II)? SOLUTION Analyze We are given the name of a complex containing only monodentate ligands, and we must determine the number of isomers the complex can form. Plan We can count the number of ligands to determine the coordination number of the Fe and then use the coordination number to predict the geometry. We can then either make a series of drawings with ligands in different positions to determine the number of isomers or deduce the number of isomers by analogy to cases we have discussed. Solve The name indicates that the complex has four carbonyl (CO) ligands and two chloro (Cl-) ligands, so its formula is Fe(CO)4Cl2. The complex therefore has a coordination number of 6, and we can assume an octahedral geometry. Like 3Co(NH3)4Cl 24 + (Figure 23.7), it has four ligands of one type and two of another. Consequently, there are two isomers possible: one with the Cl- ligands across the metal from each other, trans-[Fe(CO)4Cl2], and one with the Cl- ligands adjacent to each other, cis-[Fe(CO)4Cl2].
SECTION 23.4
Nomenclature and Isomerism in Coordination Chemistry
Comment It is easy to overestimate the number of geometric isomers. Sometimes different orientations of a single isomer are incorrectly thought to be different isomers. If two structures can be rotated so that they are equivalent, they are not isomers of each other. The problem of identifying isomers is compounded by the difficulty we often have in visualizing threedimensional molecules from their two-dimensional representations. It is sometimes easier to determine the number of isomers if we use three-dimensional models. PRACTICE EXERCISE How many isomers exist for the square-planar molecule [Pt(NH3)2ClBr]? Answer: two
The second type of stereoisomerism listed in Figure 23.19 is optical isomerism. Optical isomers, called enantiomers, are mirror images that cannot be superimposed on each other. They bear the same resemblance to each other that your left hand bears to your right hand. If you look at your left hand in a mirror, the image is identical to your right hand (쑼 FIGURE 23.22). No matter how hard you try, however, you cannot superimpose your two hands on each other. An example of a complex that exhibits this type of isomerism is the 3Co(en)343 + ion. Figure 23.22 shows the two enantiomers of this complex and their mirror-image relationship. Just as there is no way that we can twist or turn our right hand to make it look identical to our left, so also there is no way to rotate one of these enantiomers to make it identical to the other. Molecules or ions that are not superimposable on their mirror image are said to be chiral (pronounced KY-rul). Mirror
Mirror
N
N Co
Left hand
Mirror image of left hand is identical to right hand
Enantiomers of [Co(en)3]3⫹
쑿 FIGURE 23.22 Optical isomerism.
SAMPLE EXERCISE 23.6
Co
Predicting Whether a Complex Has Optical Isomers
Does either cis-3Co1en22Cl 24 + or trans-3Co(en)2Cl 24 - have optical isomers? SOLUTION Analyze We are given the chemical formula for two geometric isomers and asked to determine whether either one has optical isomers. Because en is a bidentate ligand, we know that both complexes are octahedral and both have coordination number 6. Plan We need to sketch the structures of the cis and trans isomers and their mirror images. We can draw the en ligand as two N atoms connected by an arc. If the mirror image cannot be superimposed on the original structure, the complex and its mirror image are optical isomers.
983
984
CHAPTER 23
Transition Metals and Coordination Chemistry
Solve The trans isomer of 3Co(en)2Cl 24 + and its mirror image are: Mirror
Cl
Cl
N
N
N
Co N
N Co
N
N
N
Cl
Cl
Notice that the mirror image of the isomer is identical to the original. Consequently trans3Co(en)2Cl 24 + does not exhibit optical isomerism. The mirror image of the cis isomer cannot be superimposed on the original: Mirror
N
N
N N
N
Cl
Cl Co
Co Cl
N
Cl
N N
Thus, the two cis structures are optical isomers (enantiomers). We say that cis-3Co(en)2Cl 24+ is a chiral complex. PRACTICE EXERCISE Does the square-planar complex ion 3Pt(NH3)(N3)ClBr4- have optical isomers? Explain your answer. Answer: no, because the complex is flat. This complex ion does, however, have geometric isomers (for example, the Cl and Br ligands could be cis or trans).
The properties of two optical isomers differ only if the isomers are in a chiral environment—that is, an environment in which there is a sense of right- and lefthandedness. A chiral enzyme, for example, might catalyze the reaction of one optical isomer but not the other. Consequently, one optical isomer may produce a specific physiological effect in the body, with its mirror image producing either a different effect or none at all. Chiral reactions are also extremely important in the synthesis of pharmaceuticals and other industrially important chemicals. Optical isomers are usually distinguished from each other by their interaction with plane-polarized light. If light is polarized—for example, by being passed through a sheet of polarizing film—the electric-field vector of the light is confined to a single plane (쑼 FIGURE 23.23). If the polarized light is then passed through a solution containing one optical isomer, the plane of polarization is rotated either to the right or to the left. The isomer that rotates the plane of polarization to the right is dextrorotatory; it is the
Unpolarized light
Polarizing film
Angle of rotation of plane of polarization Rotated polarized light
Light source
씰 FIGURE 23.23 Using polarized light to detect optical activity.
Polarizer axis
Polarized light
Optically active solution Analyzer
SECTION 23.5
Color and Magnetism in Coordination Chemistry
985
dextro, or d, isomer (Latin dexter, “right”). Its mirror image rotates the plane of polarization to the left; it is levorotatory and is the levo, or l, isomer (Latin laevus, “left”). The 3Co(en)343 + isomer on the right in Figure 23.22 is found experimentally to be the l isomer of this ion. Its mirror image is the d isomer. Because of their effect on plane-polarized light, chiral molecules are said to be optically active. GIVE IT SOME THOUGHT What is the similarity and what is the difference between the d and l isomers of a compound?
When a substance with optical isomers is prepared in the laboratory, the chemical environment during the synthesis is not usually chiral. Consequently, equal amounts of the two isomers are obtained, and the mixture is said to be racemic. A racemic mixture does not rotate polarized light because the rotatory effects of the two isomers cancel each other.
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23.5 COLOR AND MAGNETISM IN COORDINATION CHEMISTRY Studies of the colors and magnetic properties of transition-metal complexes have played an important role in the development of modern models for metal–ligand bonding. We discussed the various types of magnetic behavior of the transition metals in Section 23.1, and we discussed the interaction of radiant energy with matter in Section 6.3. Let’s briefly examine the significance of these two properties for transition-metal complexes before we develop a model for metal–ligand bonding.
Color In Figure 23.3 we saw the diverse range of colors seen in salts of transition-metal ions and their aqueous solutions. In general, the color of a complex depends on the identity of the metal ion, on its oxidation state, and on the ligands bound to it. 쑼 FIGURE 23.24, for instance, shows how the pale blue color characteristic of 3Cu(H2O)442 + changes to deep blue as NH3 ligands replace the H2O ligands to form 3Cu(NH3)442 + . GO FIGURE
Is the equilibrium binding constant of ammonia for Cu(II) likely to be larger or smaller than that of water for Cu(II)?
[Cu(H2O)4
]2⫹(aq)
NH3(aq)
[Cu(NH3)4
]2⫹(aq)
씱 FIGURE 23.24 The color of a coordination complex changes when the ligand changes.
986
CHAPTER 23
Transition Metals and Coordination Chemistry
Eye perceives orange since only orange light reflected
Only blue light absorbed; eye perceives orange, blue’s complementary color
650 nm
O
R
Y
750 nm 400 nm
560 nm
V 430 nm
580 nm
G B
490 nm
쑿 FIGURE 23.25 Two ways of perceiving the color orange. An object appears orange either when it reflects orange light to the eye (left), or when it transmits to the eye all colors except blue, the complement of orange (middle). Complementary colors lie opposite to each other on an artist's color wheel (right).
For a substance to have color we can see, it must absorb some portion of the spectrum of visible light. • (Section 6.1) Absorption happens, however, only if the energy needed to move an electron in the substance from its ground state to an excited state corresponds to the energy of some portion of the visible spectrum. • (Section 6.3) Thus, the particular energies of radiation a substance absorbs dictate the color we see for the substance. When an object absorbs some portion of the visible spectrum, the color we perceive is the sum of the unabsorbed portions, which are either reflected or transmitted by the object and strike our eyes. (Opaque objects reflect light, and transparent ones transmit it.) If an object absorbs all wavelengths of visible light, none reaches our eyes and the object appears black. If it absorbs no visible light, it is white if it is a solid or colorless if it is a liquid. If it absorbs all but orange light, the orange light is what reaches our eye and therefore the color we see. An interesting phenomenon of vision is that we also perceive an orange color when an object absorbs only the blue portion of the visible spectrum and all the other colors strike our eyes. This is because orange and blue are complementary colors, which means that the removal of blue from white light makes the light look orange (and the removal of orange makes the light look blue). Complementary colors can be determined with an artist’s color wheel, which shows complementary colors on opposite sides (쑿 FIGURE 23.25). SAMPLE EXERCISE 23.7
Relating Color Absorbed to Color Observed
The complex ion trans-3Co(NH3)4Cl 24 + absorbs light primarily in the red region of the visible spectrum (the most intense absorption is at 680 nm). What is the color of the complex? SOLUTION Analyze We need to relate the color absorbed by a complex (red) to the color observed for the complex. Plan For an object that absorbs only one color from the visible spectrum, the color we see is complementary to the color absorbed. We can use the color wheel of Figure 23.25 to determine the complementary color. Solve From Figure 23.25, we see that green is complementary to red, so the complex appears green. Comment As noted in Section 23.2, this green complex was one of those that helped Werner establish his theory of coordination (Table 23.3). The other geometric isomer of this complex, cis-3Co(NH3)4Cl 24 + , absorbs yellow light and therefore appears violet. PRACTICE EXERCISE A certain transition-metal complex ion absorbs at 630 nm. Which color is this ion most likely to be—blue, yellow, green, or orange? Answer: blue
The amount of light absorbed by a sample as a function of wavelength is known as the sample’s absorption spectrum. The visible absorption spectrum of a transparent sample can be determined using a spectrometer, as described in the “A Closer Look” box on page 564. The absorption spectrum of the ion 3Ti(H2O)643 + is shown in 씰 FIGURE 23.26. The absorption maximum is at 500 nm, but the graph shows that much of the yellow, green, and blue light is also absorbed. Because the sample absorbs all of these colors, what we see is the unabsorbed red and violet light, which we perceive as red-violet.
987
Crystal-field Theory
SECTION 23.6
GO FIGURE
How would this absorbance spectrum change if you decreased the concentration of the [Ti(H2O)6]3+ in solution? Blue, green, yellow absorbed; violet and red light travel to eye, solution appears red-violet
Magnetism of Coordination Compounds Many transition-metal complexes exhibit paramagnetism, as described in Sections 9.8 and 23.1. In such compounds the metal ions possess some number of unpaired electrons. It is possible to experimentally determine the number of unpaired electrons per metal ion from the measured degree of paramagnetism, and experiments reveal some interesting comparisons. Compounds of the complex ion 3Co(CN)643- have no unpaired electrons, for example, but compounds of the 3CoF643- ion have four unpaired electrons per metal ion. Both complexes contain Co(III) with a 3d6 electron configuration. • (Section 7.4) Clearly, there is a major difference in the ways in which the electrons are arranged in these two cases. Any successful bonding theory must explain this difference, and we present such a theory in the next section.
What is the electron configuration for a. the Co atom and b. the Co3 + ion? How many unpaired electrons does each possess? (See Section 7.4 to review electron configurations of ions.)
23.6 | CRYSTAL-FIELD THEORY Scientists have long recognized that many of the magnetic properties and colors of transition-metal complexes are related to the presence of d electrons in the metal cation. In this section we consider a model for bonding in transition-metal complexes, crystalfield theory, that accounts for many of the observed properties of these substances.* Because the predictions of crystal-field theory are essentially the same as those obtained with more advanced molecular-orbital theories, crystal-field theory is an excellent place to start in considering the electronic structure of coordination compounds. The ability of a metal ion to attract ligands is a Lewis acid–base interaction, in which the base—that is, the ligand—donates a pair of electrons to an empty orbital on the metal ion (씰 FIGURE 23.27). Much of the attractive interaction between the metal ion and the ligands is due, however, to the electrostatic forces between the positive charge on the metal ion and negative charges on the ligands. If the ligand is ionic, as in the case of Cl- or SCN - , the electrostatic interaction is the usual cation–anion attraction. When the ligand is a neutral molecule, as in the case of H2O or NH3, the negative ends of these polar molecules, which contain an unshared electron pair, are directed toward the metal ion. In this case, the attractive interaction is of the ion–dipole type. • (Section 11.2) In either case, the ligands are attracted strongly toward the metal ion. Because of the metal–ligand electrostatic attraction, the energy of the complex is lower than the combined energy of the separated metal ion and ligands. Although the metal ion is attracted to the ligand electrons, the metal ion’s d electrons are repulsed by the ligands. Let’s examine this effect more closely, specifically the case in which the ligands form an octahedral array around a metal ion that has coordination number 6. *The name crystal field arose because the theory was first developed to explain the properties of solid crystalline materials. The theory applies equally well to complexes in solution, however.
Absorbance
GIVE IT SOME THOUGHT
400
500 600 700 Wavelength (nm)
쑿 FIGURE 23.26 The color of [Ti(H2O)6]3ⴙ . A solution containing the 3Ti(H2O)643 + ion appears red-violet because, as its visible absorption spectrum shows, the solution does not absorb light from the violet and red ends of the spectrum. That unabsorbed light is what reaches our eyes.
n⫹
M
L
쑿 FIGURE 23.27 Metal–ligand bond formation. The ligand acts as a Lewis base by donating its nonbonding electron pair to a hybrid orbital on the metal ion. The bond that results is strongly polar with some covalent character.
988
CHAPTER 23
Transition Metals and Coordination Chemistry
GO FIGURE
Which d orbitals have lobes that point directly toward the ligands in an octahedral crystal field?
z
ⴚ ⴙ
e set (dz2, dx2⫺y2)
Energy
Free metal ion
Metal ion plus ligands (negative point charges with spherical symmetry)
ⴚ
ⴚ
ⴚ
ⴚ
t2 set (dxy, dxz, dyz)
y
x
ⴚ
ⴚ
ⴚ
ⴚ
z
dx2⫺y2
ⴚ
ⴚ
ⴚ
y
x
x
z
ⴚ
ⴚ
ⴚ ⴚ
y
ⴚ
dz2
z
In octahedral crystal field
ⴚ
ⴚ
⌬ ⴙ
z
ⴚ
ⴚ
ⴚ
ⴚ
ⴚ
y
x
ⴚ
ⴚ
ⴚ
ⴚ
ⴚ
ⴚ
ⴚ
dxy
dxz
dyz
y
x
쑿 FIGURE 23.28 Energies of d orbitals in a free metal ion, a spherically symmetric crystal field, and an octahedral crystal field.
In crystal-field theory, we consider the ligands to be negative points of charge that repel the electrons in the d orbitals of the metal ion. The energy diagram in 쑿 FIGURE 23.28 shows how these ligand point charges affect the energies of the d orbitals. First we imagine the complex as having all the ligand point charges uniformly distributed on the surface of a sphere centered on the metal ion. The average energy of the metal ion’s d orbitals is raised by the presence of this uniformly charged sphere. Hence, the energies of all five d orbitals are raised by the same amount. This energy picture is only a first approximation, however, because the ligands are not distributed uniformly on a spherical surface and, therefore, do not approach the metal ion equally from every direction. Instead, we envision the six ligands approaching along x-, y-, and z-axes, as shown on the right in Figure 23.28. This arrangement of ligands is called an octahedral crystal field. Because the metal ion’s d orbitals have different orientations and shapes, they do not all experience the same repulsion from the ligands and, therefore, do not all have the same energy under the influence of the octahedral crystal field. To see why, we must consider the shapes of the d orbitals and how their lobes are oriented relative to the ligands. Figure 23.28 shows that the lobes of the dz2 and dx2 - y2 orbitals are directed along the x-, y-, and z-axes and so point directly toward the ligand point charges. In the dxy, dxz, and dyz orbitals, however, the lobes are directed between the axes and so do not point directly toward the charges. The result of this difference in orientation—dx2 - y2 and dz2 lobes point directly toward the ligand charges; dxy, dxz, and dyz lobes do not—is that the energy of the dx2 - y2 and dz2 orbitals is higher than the energy of the dxy, dxz, and dyz orbitals. This difference in energy is represented by the red boxes in the energy diagram of Figure 23.28. It might seem like the energy of the dx2 - y2 orbital should be different from that of the dz2 orbital because the dx2 - y2 has four lobes pointing at ligands and the dz2 has only two lobes pointing at ligands. However, the dz2 orbital does have electron density in the xy plane, represented by the ring encircling the point where the two lobes meet. More advanced calculations show that two orbitals do indeed have the same energy in the presence of the octahedral crystal field. Because their lobes point directly at the negative ligand charges, electrons in the metal ion’s dz2 and dx2 - y2 orbitals experience stronger repulsions than those in the dxy, dxz, and dyz orbitals. As a result, the energy splitting shown in Figure 23.28 occurs. The three lower-energy d orbitals are called the t2 set of orbitals, and the two higher-energy ones are called the e set.* The energy gap ¢ between the two sets is often called the crystal-field splitting energy. *The labels t2 for the dxy, dxz , and dyz orbitals and e for the dz2 and dx2-y2 orbitals come from the application of a branch of mathematics called group theory to crystal-field theory. Group theory can be used to analyze the effects of symmetry on molecular properties.
SECTION 23.6
GIVE IT SOME THOUGHT Why are compounds of Ti(IV) colorless?
The magnitude of the crystal-field splitting energy and, consequently, the color of a complex depend on both the metal and the ligands. For example, 3Fe(H2O)643 + is light violet, 3Cr(H2O)643 + is a deeper violet, and 3Cr(NH3)643 + is yellow. In a ranking called the spectrochemical series, ligands are arranged in order of their abilities to increase splitting energy, as in this abbreviated list: ——Increasing ¢ ¡
Cl- 6 F- 6 H2O 6 NH3 6 en 6 NO2- (N-bonded) 6 CNThe magnitude of ¢ increases by roughly a factor of 2 from the far left to the far right of the spectrochemical series. Ligands at the low-¢ end of the spectrochemical series are termed weak-field ligands; those at the high-¢ end are termed strong-field ligands. 쑼 FIGURE 23.30 shows what happens to crystal-field splitting when the ligand is varied in a series of chromium(III) complexes. Because the Cr atom has an 3Ar43d54s1 electron configuration, Cr 3 + has the configuration 3Ar43d3 and therefore is a d3 ion. Consistent with Hund’s rule, the three 3d electrons occupy the t2 set of orbitals, with one electron in each orbital and all the spins the same. • (Section 6.8) As the crystal field exerted by the six ligands increases, ¢ increases. Because the absorption spectrum is related to this energy separation, these complexes vary in color. GO FIGURE
If you were told to add a colorless Cr(III) complex to this diagram, what would you draw and where would you place it? e e e Energy
e
t2
t2
t2
t2
[CrF6]3⫺
[Cr(H6O)6]3⫹
[Cr(NH3)6]3⫹
[Cr(CN)6]3
Green
Violet
Yellow
Yellow
쑿 FIGURE 23.30 Effect of ligand on crystal-field splitting. The greater the crystal-field strength of the ligand, the greater the energy gap Δ it causes between the t2 and e sets of the metal ion’s d orbitals.
GO FIGURE
How would you calculate the energy gap between the t2 and e orbitals from this diagram? Energy
Crystal-field theory helps us account for the colors observed in transition-metal complexes. The energy gap ¢ between the e and t2 sets of d orbitals is of the same order of magnitude as the energy of a photon of visible light. It is therefore possible for a transitionmetal complex to absorb visible light that excites an electron from a lower-energy (t2) d orbital into a higher-energy (e) one. In 3Ti(H2O)643+, for example, the Ti(III) ion has an 3Ar43d1 electron configuration. (Recall from Section 7.4 that when determining the electron configurations of transition-metal ions, we remove the s electrons first.) Ti(III) is thus called a d1 ion. In the ground state of 3Ti(H2O)643 + , the single 3d electron resides in an orbital in the t2 set (씰 FIGURE 23.29). Absorption of 495-nm light excites this electron up to an orbital in the e set, generating the absorption spectrum shown in Figure 23.26. Because this transition involves exciting an electron from one set of d orbitals to the other, we call it a d-d transition. As noted earlier, the absorption of visible radiation that produces this d-d transition causes the 3Ti(H2O)643 + ion to appear red-violet.
989
Crystal-field Theory
Light
e
of 495 nm t2
쑿 FIGURE 23.29 The d-d transition in [Ti(H2O)6]3+ is produced by the absorption of 495-nm light.
990
CHAPTER 23
Transition Metals and Coordination Chemistry
SAMPLE EXERCISE 23.8 3+
Using the Spectrochemical Series
3+
3Ti(H2O)64 , 3Ti(en)34 , and 3TiCl 643- all absorb visible light. Which one absorbs at the shortest wavelength? SOLUTION Analyze We are given three octahedral complexes, each containing Ti in the +3 oxidation state. We need to predict which complex absorbs the shortest wavelength of visible light. Plan Ti(III) is a d1 ion, so we anticipate that the absorption is due to a d-d transition in which the 3d electron is excited from a t2 orbital to an e orbital. The wavelength of the light absorbed is determined by the magnitude of the crystal-field splitting energy, ¢ . Thus, we use the position of the ligands in the spectrochemical series to predict the relative values of ¢ . The larger the splitting energy, the shorter the wavelength. Solve Of these three ligands, ethylenediamine (en) is highest in the spectrochemical series and therefore causes the largest split between the t2 and e orbitals. The larger the split, the shorter the wavelength of the light absorbed. Thus, the complex that absorbs the shortestwavelength light is 3Ti(en)343 + . PRACTICE EXERCISE The absorption spectrum of a Ti(III) complex containing the ligand L, 3TiL 643 - , shows a peak maximum at a wavelength intermediate between the wavelengths of the absorption maxima for 3TiCl 643- and 3TiF643- . What can we conclude about the place of L in the spectrochemical series? Answer: It lies between Cl- and F - .
Electron Configurations in Octahedral Complexes Crystal-field theory helps us understand the magnetic properties and some important chemical properties of transition-metal ions. From Hund’s rule, we expect electrons to always occupy the lowest-energy vacant orbitals first and to occupy a set of degenerate (same-energy) t2 orbitals one at a time with their spins parallel. • (Section 6.8) Thus, if we have a d1, d2, or d3 octahedral complex, the electrons go into the lower-energy t2 orbitals, with their spins parallel. When a fourth electron must be added, we have the two choices shown in 씱 FIGURE e 23.31: The electron can either go into an e orbital, where it will be the sole electron in the orbital, or become the second electron in a t2 orbital. Because the energy difference between the t2 and e sets is the splitting t2 energy ¢ , the energy cost of going into an e orbital rather than a t2 orbital is also ¢ . Thus, the goal of filling lowest-energy available orbitals first is met by putting the electron in a t2 orbital. There is a penalty for doing this, however, because the electron must now be paired with the electron already occupying the orbital. The difference between the energy required to pair an electron in an occupied orbital and the energy required to place that electron in an empty orbital is called the spin-pairing energy. The spin-pairing energy arises from the fact that the electrostatic repulsion between two electrons that share an orbital (and so must have opposite spins) is greater than the repulsion between two electrons that are in different orbitals and have parallel spins. In coordination complexes, the nature of the ligands and the charge on the metal ion often play major roles in determining which of the two electron arrangements shown in Figure 23.31 is used. In 3CoF643- and 3Co(CN)643-, both ligands have a 1charge. The F - ion, however, is on the low end of the spectrochemical series, so it is a weak-field ligand. The CN- ion is on the high end and so is a strong-field ligand, which means it produces a larger energy gap ¢ than the F - ion. The splittings of the d-orbital energies in these two complexes are compared in 씰 FIGURE 23.32. Cobalt(III) has an 3Ar43d6 electron configuration, so both complexes in Figure 23.32 are d6 complexes. Let’s imagine that we add these six electrons one at a time to the d orbitals of the [CoF6]3- ion. The first three go into the t2 orbitals with their spins parallel. The fourth electron could pair up in one of the t2 orbitals. The F - ion is a weak-field ligand, however, and so the energy gap ¢ between the t2 set and the e set is small. In this case, the more stable arrangement is the fourth electron in one of the e orbitals. By the e
4th e⫺ Cr 3⫹, a d 3 ion
e Or t2
4th e⫺
쑿 FIGURE 23.31 Two possibilities for adding a fourth electron to a d 3 octahedral complex. Whether the fourth electron goes into a t2 orbital or into an e orbital depends on the relative energies of the crystal-field splitting energy and the spin-pairing energy.
SECTION 23.6
GIVE IT SOME THOUGHT Are strong-field ligands more likely to lead to a high-spin complex or a low-spin complex?
In the transition metal ions of periods 5 and 6 (which have 4d and 5d valence electrons), the d orbitals are larger than in the period 4 ions (which have only 3d electrons). Thus, ions from periods 5 and 6 interact more strongly with ligands, resulting in a larger crystal-field splitting. Consequently, metal ions in periods 5 and 6 are invariably low spin in an octahedral crystal field. SAMPLE EXERCISE 23.9
Predicting the Number of Unpaired Electrons in an Octahedral Complex
Predict the number of unpaired electrons in high-spin and low-spin Fe3+ complexes that have a coordination number of 6. SOLUTION Analyze We must determine how many unpaired electrons there are in the high-spin and low-spin complexes of Fe 3 + . Plan We need to consider how the electrons populate the d orbitals in Fe 3 + when the metal is in an octahedral complex. There are two possibilities: one giving a high-spin complex and the other giving a low-spin complex. The electron configuration of Fe 3 + gives us the number of d electrons. We then determine how these electrons populate the t2 and e sets of d orbitals. In the high-spin case, the energy difference between the t2 and e orbitals is small, and the complex has the maximum number of unpaired electrons. In the low-spin case, the energy difference between the t2 and e orbitals is large, causing the t2 orbitals to be filled before any electrons occupy the e orbitals. Solve Fe 3 + is a d5 ion. In a high-spin complex, all five electrons are unpaired, with three in the t2 orbitals and two in the e orbitals. In a low-spin complex, all five electrons reside in the t2 set, so there is one unpaired electron: e High spin
e Low spin t2
t2
PRACTICE EXERCISE In octahedral complexes, for which d electron configurations is it possible to distinguish between high-spin and low-spin arrangements? Answer: d4, d5, d6, d7
Tetrahedral and Square-Planar Complexes Thus far we have considered crystal-field theory only for complexes having an octahedral geometry. When there are only four ligands in a complex, the geometry is generally tetrahedral, except for the special case of d8 metal ions, which we will discuss in a moment.
High-spin, highest number possible of unpaired electrons
Low-spin, highest number possible of paired electrons
e e
Energy
same energy argument, the fifth electron goes into the other e orbital. With all five d orbitals containing one electron, the sixth must pair up, and the energy needed to pair with a t2 electron is less than that needed to pair with an e electron. We end up with four t2 electrons and two e electrons. Figure 23.32 shows that the crystal-field splitting energy ¢ is much larger in the 3Co(CN)643- complex. In this case, the spin-pairing energy is smaller than ¢ , so the lowest-energy arrangement is the six electrons paired in the t2 orbitals. The 3CoF643- complex is a high-spin complex; that is, the electrons are arranged so that they remain unpaired as much as possible. The 3Co(CN)643- ion is a low-spin complex; that is, the electrons are arranged so that they remain paired as much as possible while still following Hund’s rule. These two electronic arrangements can be readily distinguished by measuring the magnetic properties of the complex. Experiments show that 3CoF643- has four unpaired electrons and 3Co(CN)643- has none. The absorption spectrum also shows peaks corresponding to the different values of ¢ in these two complexes.
991
Crystal-field Theory
⌬
⌬ t2
[CoF6]3⫺ t2 [Co(CN)6]
3⫺
쑿 FIGURE 23.32 High-spin and lowspin complexes. The high-spin 3CoF643- ion has a weak-field ligand and so a small ¢ value. The spin-pairing energy required to pair electrons in the t2 orbitals is greater than the e-t 2 gap energy ¢ . Therefore, filling e orbitals before any electrons are paired in t 2 orbitals is the lower-energy state. The lowspin 3Co(CN)643- ion has a strong-field ligand and so a large ¢ value. Here the spin-pairing energy is less than ¢ , making three sets of t 2-paired electrons the lower-energy state.
992
CHAPTER 23
Transition Metals and Coordination Chemistry
Energy
t2 (dxy , dyz, dxz) e (dx2⫺y2 , dz2)
쑿 FIGURE 23.33 Energies of the d orbitals in a tetrahedral crystal field. The t 2 set is the higher-energy set. Compare the relative e and t 2 energy levels here with those in Figure 23.28 for an octahedral crystal field.
GO FIGURE
Why is the dx2 - y2 orbital the highest-energy orbital in the square-planar crystal field?
Energy
dx2⫺y2
dxy
The crystal-field splitting of d orbitals in tetrahedral complexes differs from that in octahedral complexes. Four equivalent ligands can interact with a central metal ion most effectively by approaching along the vertices of a tetrahedron. It turns out—and this is not easy to explain in just a few sentences—that when d orbitals split in a tetrahedral crystal, the three t2 orbitals are raised in energy above the two e orbitals(씱 FIGURE 23.33). Because there are only four ligands instead of six, as in the octahedral case, the crystal-field splitting energy ¢ is much smaller for tetrahedral complexes. Calculations show that for the same metal ion and ligand set, ¢ for the tetrahedral complex is only four-ninths as large as for the octahedral complex. For this reason, all tetrahedral complexes are high spin; the crystal-field splitting energy is never large enough to overcome the spin-pairing energies. In a square-planar complex, four ligands are arranged about the metal ion such that all five species are in the xy plane. The resulting energy levels of the d orbitals are illustrated in 씱 FIGURE 23.34. Note in particular that the dz2 orbital is considerably lower in energy than the dx2 - y2 orbital. To understand why this is so, recall from Figure 23.28 that in an octahedral field the dz2 orbital of the metal ion interacts with the ligands positioned above and below the xy plane. There are no ligands in these two positions in a square-planar complex, which means that the dz2 orbital experiences no repulsive force and so remains in a lower-energy, more stable state. Square-planar complexes are characteristic of metal ions with a d8 electron configuration. They are nearly always low spin, with the eight d electrons spin-paired to form a diamagnetic complex. This pairing leaves the dx2 - y2 orbital empty. Such an electronic arrangement is particularly common among the ions of heavier metals, such as Pd2 + , Pt 2 + , Ir + , and Au3 + . GIVE IT SOME THOUGHT
dz2 dxz, dyz
Why is the energy of the dxz and dyz orbitals in a square-planar complex lower than that of the dxy orbital? SAMPLE EXERCISE 23.10
Square planar 쑿 FIGURE 23.34 Energies of the d orbitals in a square-planar crystal field.
Populating d Orbitals in Tetrahedral and Square-Planar Complexes
Nickel(II) complexes in which the metal coordination number is 4 can have either squareplanar or tetrahedral geometry. 3NiCl 442- is paramagnetic, and 3Ni(CN)442- is diamagnetic. One of these complexes is square planar, and the other is tetrahedral. Use the relevant crystalfield splitting diagrams in the text to determine which complex has which geometry. SOLUTION Analyze We are given two complexes containing Ni2 + and their magnetic properties. We are given two molecular geometry choices and asked to use crystal-field splitting diagrams from the text to determine which complex has which geometry. Plan We need to determine the number of d electrons in Ni2 + and then use Figure 23.33 for the tetrahedral complex and Figure 23.34 for the square-planar complex. Solve Nickel(II) has the electron configuration 3Ar43d8. Tetrahedral complexes are always high spin, and square-planar complexes are almost always low spin. Therefore, the population of the d electrons in the two geometries is dx2⫺y2 dxy
t2 (dxy, dyz, dxz) e (dx2⫺y2, dz2) Tetrahedral
dz 2 dxz, dyz
Square planar
The tetrahedral complex has two unpaired electrons, and the square-planar complex has none. We know from Section 23.1 that the tetrahedral complex must be paramagnetic and
SECTION 23.6
Crystal-field Theory
993
the square planar must be diamagnetic. Therefore, 3NiCl 442- is tetrahedral, and 3Ni(CN)442- is square planar. Comment Nickel(II) forms octahedral complexes more frequently than square-planar ones, whereas heavier d8 metals tend to favor square-planar coordination. PRACTICE EXERCISE How many unpaired electrons do you predict for the tetrahedral 3CoCl 442- ion? Answer: three
Crystal-field theory can be used to explain many observations in addition to those we have discussed. The theory is based on electrostatic interactions between ions and atoms, which essentially means ionic bonds. Many lines of evidence show, however, that
A CLOSER LOOK
In the laboratory portion of your course, you have probably seen many colorful transition-metal compounds, including those shown in 쑼 FIGURE 23.35. Many of these compounds are colored because of d-d transitions. Some colored complexes, however, including the violet permanganate ion, MnO4 -, and the yellow chromate ion, CrO42-, derive their color from a different type of excitation involving the d orbitals. The permanganate ion strongly absorbs visible light, with a maximum absorption at 565 nm. Because violet is complementary to yellow, this strong absorption in the yellow portion of the visible spectrum is responsible for the violet color of salts and solutions of the ion. What is happening during this absorption of light? The MnO4 - ion is a complex of Mn(VII). Because Mn(VII) has a d0 electron configuration, the absorption cannot be due to a d-d transition because there are no d electrons to excite! That does not mean, however, that the d orbitals are not involved in the transition. The excitation in the MnO4 - ion is due to a charge-transfer transition, in which an electron on one oxygen ligand is excited into a vacant d orbital on the Mn ion (씰 FIGURE 23.36). In essence, an electron is transferred from a ligand to the metal, so this transition is called a ligand-to-metal charge-transfer (LMCT) transition. An LMCT transition is also responsible for the color of the CrO42- , which is a d0 Cr(VI) complex. Also shown in Figure 23.35 is a salt of the perchlorate ion (ClO4 -). Like MnO4 - , ClO4 - is tetrahedral and has its central atom
KMnO4
in the +7 oxidation state. However, because the Cl atom does not have low-lying d orbitals, exciting a Cl electron requires a more energetic photon than does MnO4 - . The first absorption for ClO4 - is in the ultraviolet portion of the spectrum, so all the visible light is transmitted and the salt appears white. Other complexes exhibit charge-transfer excitations in which an electron from the metal atom is excited to an empty orbital on a ligand. Such an excitation is called a metal-to-ligand charge-transfer (MLCT) transition. Charge-transfer transitions are generally more intense than d-d transitions. Many metal-containing pigments used for oil painting, such as cadmium yellow (CdS), chrome yellow (PbCrO4), and red ochre (Fe2O3), have intense colors because of charge-transfer transitions. RELATED EXERCISES: 23.82, 23.83
K2CrO4
Empty Mn 3d orbitals t2 set Energy
CHARGE-TRANSFER COLOR
e set
Filled ligand orbitals
씱 FIGURE 23.36 Ligand-tometal charge-transfer transition in MnO4 ⴚ . As shown by the blue arrow, an electron is excited from a nonbonding pair on O into one of the empty d orbitals on Mn.
KClO4
쑿 FIGURE 23.35 The colors of compounds can arise from charge-transfer transitions. KMnO4 and K2CrO4 are colored due to ligandto-metal charge-transfer transitions in their anions. KClO4’s anion has no occupied d orbitals and its charge-transfer transition is at higher energy, corresponding to ultraviolet absorption; therefore it appears white.
994
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Transition Metals and Coordination Chemistry
the bonding in complexes must have some covalent character. Therefore, molecularorbital theory • (Sections 9.7 and 9.8) can also be used to describe the bonding in complexes, although the application of molecular-orbital theory to coordination compounds is beyond the scope of our discussion. Crystal-field theory, although not entirely accurate in all details, provides an adequate and useful first description of the electronic structure of complexes. SAMPLE INTEGRATIVE EXERCISE
Putting Concepts Together
The oxalate ion has the Lewis structure shown in Table 23.4. (a) Show the geometry of the complex formed when this ion complexes with cobalt(II) to form [Co(C2O4)(H2O)4]. (b) Write the formula for the salt formed when three oxalate ions complex with Co(II), assuming that the charge-balancing cation is Na + . (c) Sketch all the possible geometric isomers for the cobalt complex formed in part (b). Are any of these isomers chiral? Explain. (d) The equilibrium constant for the formation of the cobalt(II) complex produced by coordination of three oxalate anions, as in part (b), is 5.0 * 109, and the equilibrium constant for formation of the cobalt(II) complex with three molecules of ortho-phenanthroline (Table 23.4) is 9 * 1019. From these results, what conclusions can you draw regarding the relative Lewis base properties of the two ligands toward cobalt(II)? (e) Using the approach described in Sample Exercise 17.14, calculate the concentration of free aqueous Co(II) ion in a solution initially containing 0.040 M oxalate (aq) and 0.0010 M Co2 + (aq). SOLUTION (a) The complex formed by coordination of one oxalate ion is octahedral:
(b) Because the oxalate ion has a charge of 2 - , the net charge of a complex with three oxalate anions and one Co2 + ion is 4 - . Therefore, the coordination compound has the formula Na4[Co(C2O4)3]. (c) There is only one geometric isomer. The complex is chiral, however, in the same way the 3Co(en)343 + complex is chiral (Figure 23.22). The two mirror images are not superimposable, so there are two enantiomers: 4⫺
4⫺
(d) The ortho-phenanthroline ligand is bidentate, like the oxalate ligand, so they both exhibit the chelate effect. Thus, we conclude that ortho-phenanthroline is a stronger Lewis base toward Co2 + than oxalate. This conclusion is consistent with what we learned about bases in Section 16.7, that nitrogen bases are generally stronger than oxygen bases. (Recall, for example, that NH3 is a stronger base than H2O.) (e) The equilibrium we must consider involves 3 mol of oxalate ion (represented as Ox 2- ). Co2 + (aq) + 3 Ox 2-(aq) Δ 3Co(Ox)344-(aq)
Chapter Summary and Key Terms
995
The formation-constant expression is Kf =
33Co(Ox)344-4 3Co2 + 43Ox 2-43
Because Kf is so large, we can assume that essentially all the Co2 + is converted to the oxalato complex. Under that assumption, the final concentration of 3Co(Ox)344- is 0.0010 M and that of oxalate ion is 3Ox 2-4 = (0.040) - 3(0.0010) = 0.037 M (three Ox 2- ions react with each Co2 + ion). We then have 3Co2 + 4 = xM, 3Ox 2-4 ⬵ 0.037 M, 33Co(Ox)344-4 ⬵ 0.0010 M Inserting these values into the equilibrium-constant expression, we have Kf =
(0.0010) x(0.037)3
= 5 * 109
Solving for x, we obtain 4 * 10-9 M. From this, we see that the oxalate has complexed all but a tiny fraction of the Co2 + in solution.
CHAPTER SUMMARY AND KEY TERMS Metallic elements occur in nature in minerals, which are solid inorganic compounds found in nature. Metallurgy is the science and technology of extracting metals from the earth and processing them for further use. Transition metals are characterized by incomplete filling of the d orbitals. The presence of d electrons in transition elements leads to multiple oxidation states. As we proceed through the transition metals in a given row of the periodic table, the attraction between the nucleus and the valence electrons increases more markedly for d electrons than for s electrons. As a result, the later transition elements in a period tend to have lower oxidation states. The atomic and ionic radii of period 5 transition metals are larger than those of period 4 metals. The transition metals of periods 5 and 6 have comparable atomic and ionic radii and are also similar in other properties. This similarity is due to the lanthanide contraction. The presence of unpaired electrons in valence orbitals leads to magnetic behavior in transition metals and their compounds. In ferromagnetic, ferrimagnetic, and antiferromagnetic substances the unpaired electron spins on atoms in a solid are affected by spins on neighboring atoms. In a ferromagnetic substance the spins all point in the same direction. In an antiferromagnetic substance the spins point in opposite directions and cancel one another. In a ferrimagnetic substance the spins point in opposite directions but do not fully cancel. Ferromagnetic and ferrimagnetic substances are used to make permanent magnets. SECTION 23.1.
Coordination compounds are substances that contain metal complexes. Metal complexes contain metal ions bonded to several surrounding anions or molecules known as ligands. The metal ion and its ligands make up the coordination sphere of the complex. The number of atoms attached to the metal ion is the coordination number of the metal ion. The most common coordination numbers are 4 and 6; the most common coordination geometries are tetrahedral, square planar, and octahedral.
SECTION 23.2
SECTION 23.3 Ligands that occupy only one site in a coordination sphere are called monodentate ligands. The atom of the ligand that bonds to the metal ion is the donor atom. Ligands that have two donor atoms are bidentate ligands. Polydentate ligands have three or more
donor atoms. Bidentate and polydendate ligands are also called chelating agents. In general, chelating agents form more stable complexes than do related monodentate ligands, an observation known as the chelate effect. Many biologically important molecules, such as the porphyrins, are complexes of chelating agents. A related group of plant pigments known as chlorophylls is important in photosynthesis, the process by which plants use solar energy to convert CO2 and H2O into carbohydrates. SECTION 23.4 In naming coordination compounds, the number and type of ligands attached to the metal ion are specified, as is the oxidation state of the metal ion. Isomers are compounds with the same composition but different arrangements of atoms and therefore different properties. Structural isomers differ in the bonding arrangements of the ligands. Linkage isomerism occurs when a ligand can coordinate to a metal ion through either of two donor atoms. Coordinationsphere isomers contain different ligands in the coordination sphere. Stereoisomers are isomers with the same chemical bonding arrangements but different spatial arrangements of ligands. The most common forms of stereoisomerism are geometric isomerism and optical isomerism. Geometric isomers differ from one another in the relative locations of donor atoms in the coordination sphere; the most common are cis-trans isomers. Optical isomers are nonsuperimposable mirror images of each other. Geometric isomers differ from one another in their chemical and physical properties; optical isomers, or enantiomers, are chiral, however, meaning that they have a specific “handedness” and differ only in the presence of a chiral environment. Optical isomers can be distinguished from one another by their interactions with plane-polarized light; solutions of one isomer rotate the plane of polarization to the right (dextrorotatory), and solutions of its mirror image rotate the plane to the left (levorotatory). Chiral molecules, therefore, are optically active. A 50 –50 mixture of two optical isomers does not rotate plane-polarized light and is said to be racemic.
A substance has a particular color because it either reflects or transmits light of that color or absorbs light of the complementary color. The amount of light absorbed by a sample as a SECTION 23.5
996
CHAPTER 23
Transition Metals and Coordination Chemistry
function of wavelength is known as its absorption spectrum. The light absorbed provides the energy to excite electrons to higher-energy states. It is possible to determine the number of unpaired electrons in a complex from its degree of paramagnetism. Compounds with no unpaired electrons are diamagnetic. SECTION 23.6 Crystal-field theory successfully accounts for many properties of coordination compounds, including their color and magnetism. In crystal-field theory, the interaction between metal ion and ligand is viewed as electrostatic. Because some d orbitals point right at the ligands whereas others point between them, the ligands split the energies of the metal d orbitals. For an octahedral complex, the d orbitals are split into a lower-energy set of three degenerate orbitals (the t2 set) and a higher-energy set of two degenerate orbitals (the e set). Visible light can cause a d-d transition, in which an electron is excited from a lower-energy d orbital to a higher-energy d orbital. The
spectrochemical series lists ligands in order of their ability to in-
crease the split in d-orbital energies in octahedral complexes. Strong-field ligands create a splitting of d-orbital energies that is large enough to overcome the spin-pairing energy. The d electrons then preferentially pair up in the lower-energy orbitals, producing a low-spin complex. When the ligands exert a weak crystal field, the splitting of the d orbitals is small. The electrons then occupy the higher-energy d orbitals in preference to pairing up in the lowerenergy set, producing a high-spin complex. Crystal-field theory also applies to tetrahedral and square-planar complexes, which leads to different d-orbital splitting patterns. In a tetrahedral crystal field, the splitting of the d orbitals results in a higher-energy t2 set and a lower-energy e set, the opposite of the octahedral case. The splitting by a tetrahedral crystal field is much smaller than that by an octahedral crystal field, so tetrahedral complexes are always high-spin complexes.
KEY SKILLS • Describe the periodic trends in radii and oxidation states of the transition-metal ions, including the origin and effect of the lanthanide contraction. (Section 23.1) • Determine the oxidation number and number of d electrons for metal ions in complexes. (Section 23.2) • Distinguish between chelating and nonchelating ligands. (Section 23.3) • Name coordination compounds given their formula and write their formula given their name. (Section 23.4) • Recognize and draw the geometric isomers of a complex. (Section 23.4) • Recognize and draw the optical isomers of a complex. (Section 23.4) • Use crystal-field theory to explain the colors and to determine the number of unpaired electrons in a complex. (Sections 23.5 and 23.6)
EXERCISES VISUALIZING CONCEPTS 23.1 This chart shows the variation in an important property of the metals from K through Ge. Is the property atomic radius, electronegativity, or first ionization energy? Explain your choice. [Section 23.1] K
Ca
Sc
Ti
V
Cr Mn Fe
Co Ni Cu Zn Ga Ge
23.2 (a) Draw the structure for Pt(en)Cl2. (b) What is the coordination number for platinum in this complex, and what is the coordination geometry? (c) What is the oxidation state of the platinum? [Section 23.2] 23.3 Draw the Lewis structure for the ligand shown in the next column. (a) Which atoms can serve as donor atoms? Classify this ligand as monodentate, bidentate, or tridentate. (b) How many of these ligands are needed to fill the coordination sphere in an octahedral complex? [Section 23.2]
NH2CH2CH2NHCH2CO2⫺ 23.4 The complex ion shown here has a 1 - charge. Name the complex ion. [Section 23.4] ⫽N ⫽ Cl ⫽H ⫽ Pt
997
Exercises 23.5 There are two geometric isomers of octahedral complexes of the type MA3X3, where M is a metal and A and X are monodentate ligands. Of the complexes shown here, which are identical to (1) and which are the geometric isomers of (1)? [Section 23.4]
(1)
(2)
(3)
(4)
(5)
23.6 Which of the complexes shown here are chiral? Explain. [Section 23.4] ⫽ Cr
⫽ NH2CH2CH2NH2
⫽ Cl
⫽ NH3
23.8 Which of these crystal-field splitting diagrams represents: (a) a weak-field octahedral complex of Fe 3 + , (b) a strong-field octahedral complex of Fe 3 + , (c) a tetrahedral complex of Fe 3 + , (d) a tetrahedral complex of Ni2 + ? (The diagrams do not indicate the relative magnitudes of ¢ .) [Section 23.6]
(1)
(2)
(3)
(4)
23.9 In the linear crystal field shown here, the negative charges are on the z-axis. Using Figure 23.28 as a guide, predict which d orbital has lobes closest to the charges. Which two have lobes farthest from the charges? Predict the crystal-field splitting of the d orbitals in linear complexes. [Section 23.6] z
ⴚ y (1)
(2)
(3)
(4)
ⴙ
23.7 The solutions shown here each have an absorption spectrum with a single absorption peak like that shown in Figure 23.26. What color does each solution absorb most strongly? [Section 23.5]
x
ⴚ 23.10 Two Fe(II) complexes are both low spin but have different ligands. A solution of one is green and a solution of the other is red. Which solution is likely to contain the complex that has the stronger-field ligand? [Section 23.6]
THE TRANSITION METALS (section 23.1) 23.11 Explain the lanthanide contraction, and describe how it affects the properties of the transition-metal elements. 23.12 Sketch a plot of atomic radius versus number of valence d electrons for the period 5 transition metals, and explain the trend. 23.13 The + 2 oxidation state is common for almost all the transition metals. Suggest an explanation. 23.14 No compounds are known in which scandium is in the +2 oxidation state. Suggest an explanation. 23.15 Write out the ground-state electron configurations of (a) Ti3+, (b) Ru2 + , (c) Au3 + , (d) Mn4 + . 23.16 How many electrons are in the valence d orbitals in these transition-metal ions? (a) Co3 + , (b) Cu + , (c) Cd2 + , (d) Os3 + . 23.17 Explain the difference between a diamagnetic substance and a paramagnetic substance.
23.18 Distinguish among a ferromagnetic substance, an antiferromagnetic substance, and a ferrimagnetic substance. 23.19 What kind of magnetism is exhibited by this diagram:
Magnetic field Put in vertical magnetic field 23.20 The most important oxides of iron are magnetite, Fe 3O4, and hematite, Fe 2O3. (a) What are the oxidation states of iron in these compounds? (b) One of these iron oxides is ferrimagnetic, and the other is antiferromagnetic. Which iron oxide is likely to show which type of magnetism? Explain.
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TRANSITION-METAL COMPLEXES (section 23.2) 23.21 (a) What is the difference between Werner’s concepts of primary valence and secondary valence? What terms do we now use for these concepts? (b) Why can the NH3 molecule serve as a ligand but the BH3 molecule cannot?
the metal? (c) Suppose you experimentally determine that this complex exists in aqueous solution as a single species. Suggest a likely coordination number and the number and type of each ligand.
23.22 (a) What is the meaning of the term coordination number as it applies to metal complexes? (b) Give an example of a ligand that is neutral and one that is negatively charged. (c) Would you expect ligands that are positively charged to be common? Explain. (d) What type of chemical bonding is characteristic of coordination compounds? Illustrate with the compound Co(NH3)6Cl3. (e) What are the most common coordination numbers for metal complexes?
23.25 Indicate the coordination number of the metal and the oxidation number of the metal as well as the number and type of each donor atom of the ligands for each of the following complexes: (a) Na2[CdCl4] (b) K2[MoOCl4] (c) [Co(NH3)4Cl2]Cl (d) 3Ni(CN)543(e) K3[V(C2O4)3] (f) [Zn(en)2]Br2
23.23 A complex is written as NiBr2 # 6 NH3. (a) What is the oxidation state of the Ni atom in this complex? (b) What is the likely coordination number for the complex? (c) If the complex is treated with excess AgNO3(aq), how many moles of AgBr will precipitate per mole of complex?
23.24 A certain complex of metal M is formulated as MCl 3 # 3 H2O. The coordination number of the complex is not known but is expected to be 4 or 6. (a) Would conductivity measurements provide information about the coordination number? (b) In using conductivity measurements to test which ligands are bound to the metal ion, what assumption is made about the rate at which ligands enter or leave the coordination sphere of
23.26 Indicate the coordination number of the metal and the oxidation number of the metal as well as the number and type of each donor atom of the ligands for each of the following complexes: (a) K3[Co(CN)6] (b) Na2[CdBr4] (c) [Pt(en)3](ClO4)4 (d) 3Co(en)2(C2O4)4 + (e) NH4[Cr(NH3)2(NCS)4] (f) [Cu(bipy)2I]I
COMMON LIGANDS IN COORDINATION CHEMISTRY (section 23.3) 23.27 (a) What is the difference between a monodentate ligand and a bidentate ligand? (b) How many bidentate ligands are necessary to fill the coordination sphere of a six-coordinate complex? (c) You are told that a certain molecule can serve as a tridentate ligand. Based on this statement, what do you know about the molecule? 23.28 For each of the following polydentate ligands, determine (i) the maximum number of coordination sites that the ligand can occupy on a single metal ion and (ii) the number and type of donor atoms in the ligand: (a) ethylenediamine (en), (b) bipyridine (bipy), (c) the oxalate anion (C2O42-), (d) the 2ion of the porphine molecule (Figure 23.13); (e) 3EDTA44- . 23.29 Polydentate ligands can vary in the number of coordination positions they occupy. In each of the following, identify the polydentate ligand present and indicate the probable number of coordination positions it occupies: (a) [Co(NH3)4(o-phen)]Cl3 (b) [Cr(C2O4)(H2O)4]Br (c) 3Cr(EDTA)(H2O)4(d) [Zn(en)2](ClO4)2 23.30 Indicate the likely coordination number of the metal in each of the following complexes: (a) [Rh(bipy)3](NO3)3 (b) Na3[Co(C2O4)2Cl2] (c) [Cr(o-phen)3](CH3COO)3 (d) Na2[Co(EDTA)Br] 23.31 (a) What is meant by the term chelate effect? (b) What thermodynamic factor is generally responsible for the chelate effect? (c) Why are polydentate ligands often called sequestering agents?
23.32 Pyridine (C5H5N), abbreviated py, is the molecule N (a) Why is pyridine referred to as a monodentate ligand? (b) For the equilibrium reaction 3Ru(py)4(bipy)42 + + 2 py Δ 3Ru(py)642 + + bipy what would you predict for the magnitude of the equilibrium constant? Explain your answer. 23.33 Is the following ligand a chelating one? Explain. N
N 23.34 What is the geometry about the metal center in this complex? Would you expect this complex to have counterions? Explain.
N
N M
N N
Exercises
999
NOMENCLATURE AND ISOMERISM IN COORDINATION CHEMISTRY (section 23.4) 23.35 Write the formula for each of the following compounds, being sure to use brackets to indicate the coordination sphere: (a) hexaamminechromium(III) nitrate (b) tetraamminecarbonatocobalt(III) sulfate (c) dichlorobis(ethylenediamine)platinum(IV) bromide (d) potassium diaquatetrabromovanadate(III) (e) bis(ethylenediamine)zinc(II) tetraiodomercurate(II)
23.39 By writing formulas or drawing structures related to any one of these three complexes, [Co(NH3)4Br2]Cl [Pd(NH3)2(ONO)2] cis-3V(en)2Cl 24 + ,
23.36 Write the formula for each of the following compounds, being sure to use brackets to indicate the coordination sphere: (a) tetraaquadibromomanganese(III) perchlorate (b) bis(bipyridyl)cadmium(II) chloride (c) potassium tetrabromo(ortho-phenanthroline)cobaltate (III) (d) cesium diamminetetracyanochromate(III) (e) tris(ethylenediammine)rhodium(III) tris(oxalato)cobaltate(III)
23.40 (a) Draw the two linkage isomers of 3Co(NH3)5SCN42 + . (b) Draw the two geometric isomers of 3Co(NH3)3Cl 342 + . (c) Two compounds with the formula Co(NH3)5ClBr can be prepared. Use structural formulas to show how they differ. What kind of isomerism does this illustrate?
23.37 Write the names of the following compounds, using the standard nomenclature rules for coordination complexes: (a) [Rh(NH3)4Cl2]Cl (b) K2[TiCl6] (c) MoOCl4 (d) [Pt(H2O)4(C2O4)]Br2 23.38 Write names for the following coordination compounds: (a) [Cd(en)Cl2] (b) K4[Mn(CN)6] (c) [Cr(NH3)5CO3]Cl (d) [Ir(NH3)4(H2O)2](NO3)3
illustrate (a) geometric isomerism, (b) linkage isomerism, (c) optical isomerism, (d) coordination-sphere isomerism.
23.41 A four-coordinate complex MA2B2 is prepared and found to have two different isomers. Is it possible to determine from this information whether the complex is square planar or tetrahedral? If so, which is it? 23.42 Consider an octahedral complex MA3B3. How many geometric isomers are expected for this compound? Will any of the isomers be optically active? If so, which ones? 23.43 Sketch all the possible stereoisomers of (a) tetrahedral [Cd(H2O)2Cl2], (b) square-planar 3IrCl 2(PH3)24- , (c) octahedral 3Fe(o-phen)2Cl 24 + . 23.44 Sketch all the possible stereoisomers of (a) 3Rh(bipy)(o-phen)243 + , (b) 3Co(NH3)3(bipy)Br42 + , (c) square-planar [Pd(en)(CN)2].
COLOR AND MAGNETISM IN COORDINATION CHEMISTRY; CRYSTAL-FIELD THEORY (sections 23.5 and 23.6) 23.45 (a) Can we see light that is 300 nm in wavelength? 500 nm in wavelength? (b) What is meant by the term complementary color? (c) What is the significance of complementary colors in understanding the colors of metal complexes? (d) If a complex absorbs light at 610 nm, what is the energy of this absorption in kJ>mol? 23.46 (a) A complex absorbs light in the range of 200–300 nm. Do you expect it to have visible color? (b) A solution of a compound appears green. Does this observation necessarily mean that all colors of visible light other than green are absorbed by the solution? Explain. (c) What information is usually presented in a visible absorption spectrum of a compound? (d) What energy is associated with an absorption at 440 nm in kJ>mol? 23.47 Is it possible for a low-spin octahedral Fe(II) complex to be paramagnetic? Explain. 23.48 If a transition-metal complex has an even number of valence d electrons, does it necessarily mean that the complex is diamagnetic? Explain. 23.49 In crystal-field theory, ligands are modeled as if they are point negative charges. What is the basis of this assumption, and how does it relate to the nature of metal–ligand bonds? 23.50 Explain why the dxy, dxz, and dyz orbitals lie lower in energy than the dz2 and dx2-y2 orbitals in the presence of an octahedral arrangement of ligands about the central metal ion.
23.51 (a) Sketch a diagram that shows the definition of the crystal-field splitting energy (¢) for an octahedral crystal field. (b) What is the relationship between the magnitude of ¢ and the energy of the d-d transition for a d1 complex? (c) Calculate ¢ in kJ>mol if a d1 complex has an absorption maximum at 545 nm. 23.52 As shown in Figure 23.26, the d-d transition of 3Ti(H2O)643 + produces an absorption maximum at a wavelength of about 500 nm. (a) What is the magnitude of ¢ for 3Ti(H2O)643 + in kJ>mol? (b) What is the spectrochemical series? How would the magnitude of ¢ change if the H2O ligands in 3Ti(H2O)643 + were replaced with NH3 ligands? 23.53 Explain why many cyano complexes of divalent transitionmetal ions are yellow, whereas many aqua complexes of these ions are blue or green. 23.54 The 3Ni(H2O)642 + ion has an absorption maximum at about 725 nm, whereas the 3Ni(NH3)642 + ion absorbs at about 570 nm. Predict the color of a solution of each ion. (b) The 3Ni(en)342 + ion absorption maximum occurs at about 545 nm, and that of the 3Ni(bipy)342 + ion occurs at about 520 nm. From these data, indicate the relative strengths of the ligand fields created by the four ligands involved. 23.55 Give the number of (valence) d electrons associated with the central metal ion in each of the following complexes: (a) K3[TiCl6], (b) Na3[Co(NO2)6], (c) [Ru(en)3]Br3, (d) [Mo(EDTA)]ClO4, (e) K3[ReCl6].
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23.56 Give the number of (valence) d electrons associated with the central metal ion in each of the following complexes: (a) K3[Fe(CN)6], (b) [Mn(H2O)6](NO3)2, (c) Na[Ag(CN)2], (d) [Cr(NH3)4Br2]ClO4, (e) 3Sr(EDTA)42- . 23.57 A classmate says, “A weak-field ligand usually means the complex is high spin.” Is your classmate correct? Explain. 23.58 A classmate says, “A strong-field ligand means that the ligand binds strongly to the metal ion.” Is your classmate correct? Explain. 23.59 For each of the following metals, write the electronic configuration of the atom and its 2 + ion: (a) Mn, (b) Ru, (c) Rh. Draw the crystal-field energy-level diagram for the d orbitals of an octahedral complex, and show the placement of the d electrons for each 2 + ion, assuming a strong-field complex. How many unpaired electrons are there in each case? 23.60 For each of the following metals, write the electronic configuration of the atom and its 3 + ion: (a) Ru, (b) Mo, (c) Co. Draw the crystal-field energy-level diagram for the d orbitals of an octahedral complex, and show the placement of the d electrons for each 3 + ion, assuming a weak-field complex. How many unpaired electrons are there in each case?
23.61 Draw the crystal-field energy-level diagrams and show the placement of d electrons for each of the following: (a) 3Cr(H2O)642 + (four unpaired electrons), (b) 3Mn(H2O)642 + (high spin), (c) 3Ru(NH3)5H2O42 + (low spin), (d) 3IrCl 642- (low spin), (e) 3Cr(en)343 + , (f) 3NiF644-. 23.62 Draw the crystal-field energy-level diagrams and show the placement of electrons for the following complexes: (a) 3VCl 643-, (b) 3FeF643- (a high-spin complex), (c) 3Ru(bipy)343 + (a lowspin complex), (d) 3NiCl 442- (tetrahedral), (e) 3PtBr642-, (f) 3Ti(en)342 + . 23.63 The complex 3Mn(NH3)642 + contains five unpaired electrons. Sketch the energy-level diagram for the d orbitals, and indicate the placement of electrons for this complex ion. Is the ion a high-spin or a low-spin complex? 23.64 The ion 3Fe(CN)643- has one unpaired electron, whereas 3Fe(NCS)643- has five unpaired electrons. From these results, what can you conclude about whether each complex is high spin or low spin? What can you say about the placement of NCS- in the spectrochemical series?
ADDITIONAL EXERCISES 23.65 The Curie temperature is the temperature at which a ferromagnetic solid switches from ferromagnetic to paramagnetic, and for nickel, the Curie temperature is 354 °C. Knowing this, you tie a string to two paper clips made of nickel and hold the paper clips near a permanent magnet. The magnet attracts the paper clips, as shown in the photograph on the left. Now you heat one of the paper clips with a cigarette lighter, and the clip drops (right photograph). Explain what happened.
Complex
Molar Conductance (ohm-1)* of 0.050 M Solution
Pt(NH3)6Cl4 Pt(NH3)4Cl4 Pt(NH3)3Cl4 Pt(NH3)2Cl4 KPt(NH3)Cl5
523 228 97 0 108
*The ohm is a unit of resistance; conductance is the inverse of resistance.
23.66 Explain why the transition metals in periods 5 and 6 have nearly identical radii in each group. 23.67 Based on the molar conductance values listed here for the series of platinum(IV) complexes, write the formula for each complex so as to show which ligands are in the coordination sphere of the metal. By way of example, the molar conductances of 0.050 M NaCl and BaCl2 are 107 ohm-1 and 197 ohm-1, respectively.
23.68 (a) A compound with formula RuCl 3 # 5 H2O is dissolved in water, forming a solution that is approximately the same color as the solid. Immediately after forming the solution, the addition of excess AgNO3(aq) forms 2 mol of solid AgCl per mole of complex. Write the formula for the compound, showing which ligands are likely to be present in the coordination sphere. (b) After a solution of RuCl 3 # 5 H2O has stood for about a year, addition of AgNO3(aq) precipitates 3 mol of AgCl per mole of complex. What has happened in the ensuing time? 23.69 Sketch the structure of the complex in each of the following compounds and give the full compound name: (a) cis-[Co(NH3)4(H2O)2](NO3)2 (b) Na2[Ru(H2O)Cl5] (c) trans-NH4[Co(C2O4)2(H2O)2] (d) cis-[Ru(en)2Cl2] 23.70 (a) Which complex ions in Exercise 23.69 have a mirror plane? (b) Will any of the complexes be optically active? Explain. 23.71 The molecule dimethylphosphinoethane [(CH3)2PCH2CH2P(CH3)2, which is abbreviated dmpe] is used as a ligand for some complexes that serve as catalysts. A complex that contains this ligand is Mo(CO)4(dmpe). (a) Draw the Lewis structure for dmpe, and compare it with ethylenediammine as a coordinating ligand. (b) What is the oxidation state of Mo in Na2[Mo(CN)2(CO)2(dmpe)]? (c) Sketch the structure of the 3Mo(CN)2(CO)2(dmpe)42- ion, including all the possible isomers.
Additional Exercises 23.72 Although the cis configuration is known for [Pt(en)Cl2], no trans form is known. (a) Explain why the trans compound is not possible. (b) Would NH2CH2CH2CH2CH2NH2 be more likely than en (NH2CH2CH2NH2) to form the trans compound? Explain. 23.73 The acetylacetone ion forms very stable complexes with many metallic ions. It acts as a bidentate ligand, coordinating to the metal at two adjacent positions. Suppose that one of the CH3 groups of the ligand is replaced by a CF3 group, as shown here, ⫺
H
Trifluoromethyl acetylacetonate CF 3 (tfac)
C C
C
O
O
CH3
Sketch all possible isomers for the complex with three tfac ligands on cobalt(III). (You can use the symbol to represent the ligand.) 23.74 Give brief statements about the relevance of the following complexes in living systems: (a) hemoglobin, (b) chlorophylls, (c) siderophores. 23.75 Write balanced chemical equations to represent the following observations. (In some instances the complex involved has been discussed previously in the text.) (a) Solid silver chloride dissolves in an excess of aqueous ammonia. (b) The green complex [Cr(en)2Cl2]Cl, on treatment with water over a long time, converts to a brown-orange complex. Reaction of AgNO3 with a solution of the product precipitates 3 mol of AgCl per mole of Cr present. (Write two chemical equations.) (c) When an NaOH solution is added to a solution of Zn(NO3)2, a precipitate forms. Addition of excess NaOH solution causes the precipitate to dissolve. (Write two chemical equations.) (d) A pink solution of Co(NO3)2 turns deep blue on addition of concentrated hydrochloric acid. 23.76 Some metal complexes have a coordination number of 5. One such complex is Fe(CO)5, which adopts a trigonal bipyramidal geometry (see Figure 9.8). (a) Write the name for Fe(CO)5, using the nomenclature rules for coordination compounds. (b) What is the oxidation state of Fe in this compound? (c) Suppose one of the CO ligands is replaced with a CN- ligand, forming 3Fe(CO)4 (CN)4- . How many geometric isomers would you predict this complex could have? 23.77 Which of the following objects is chiral: (a) a left shoe, (b) a slice of bread, (c) a wood screw, (d) a molecular model of Zn(en)Cl2, (e) a typical golf club? 23.78 The complexes 3V(H2O)643 + and 3VF643- are both known. (a) Draw the d-orbital energy-level diagram for V(III) octahedral complexes. (b) What gives rise to the colors of these complexes? (c) Which of the two complexes would you expect to absorb light of higher energy? Explain. [23.79] One of the more famous species in coordination chemistry is the Creutz–Taube complex, 5⫹
(NH3)5RuN
NRu(NH3)5
1001
It is named for the two scientists who discovered it and initially studied its properties. The central ligand is pyrazine, a planar six-membered ring with nitrogens at opposite sides. (a) How can you account for the fact that the complex, which has only neutral ligands, has an odd overall charge? (b) The metal is in a low-spin configuration in both cases. Assuming octahedral coordination, draw the d-orbital energy-level diagram for each metal. (c) In many experiments the two metal ions appear to be in exactly equivalent states. Can you think of a reason that this might appear to be so, recognizing that electrons move very rapidly compared to nuclei? 23.80 Solutions of 3Co(NH3)642 + , 3Co(H2O)642 + (both octahedral), and 3CoCl 442- (tetrahedral) are colored. One is pink, one is blue, and one is yellow. Based on the spectrochemical series and remembering that the energy splitting in tetrahedral complexes is normally much less than that in octahedral ones, assign a color to each complex. 23.81 Oxyhemoglobin, with an O2 bound to iron, is a low-spin Fe(II) complex; deoxyhemoglobin, without the O2 molecule, is a high-spin complex. (a) Assuming that the coordination environment about the metal is octahedral, how many unpaired electrons are centered on the metal ion in each case? (b) What ligand is coordinated to the iron in place of O2 in deoxyhemoglobin? (c) Explain in a general way why the two forms of hemoglobin have different colors (hemoglobin is red, whereas deoxyhemoglobin has a bluish cast). (d) A 15-minute exposure to air containing 400 ppm of CO causes about 10% of the hemoglobin in the blood to be converted into the carbon monoxide complex, called carboxyhemoglobin. What does this suggest about the relative equilibrium constants for binding of carbon monoxide and O2 to hemoglobin? (e) CO is a strong-field ligand. What color might you expect carboxyhemoglobin to be? [23.82] Consider the tetrahedral anions VO43- (orthovanadate ion), CrO42- (chromate ion), and MnO4 - (permanganate ion). (a) These anions are isoelectronic. What does this statement mean? (b) Would you expect these anions to exhibit d-d transitions? Explain. (c) As mentioned in “A Closer Look” on charge-transfer color, the violet color of MnO4 - is due to a ligand-to-metal charge transfer (LMCT) transition. What is meant by this term? (d) The LMCT transition in MnO4 - occurs at a wavelength of 565 nm. The CrO42- ion is yellow. Is the wavelength of the LMCT transition for chromate larger or smaller than that for MnO4 -? Explain. (e) The VO43- ion is colorless. Do you expect the light absorbed by the LMCT to fall in the UV or the IR region of the electromagnetic spectrum? Explain your reasoning. [23.83] Given the colors observed for VO43- (orthovanadate ion), CrO42- (chromate ion), and MnO4 - (permanganate ion) (see Exercise 23.82), what can you say about how the energy separation between the ligand orbitals and the empty d orbitals changes as a function of the oxidation state of the transition metal at the center of the tetrahedral anion? [23.84] The red color of ruby is due to the presence of Cr(III) ions at octahedral sites in the close-packed oxide lattice of Al2O3. Draw the crystal-field splitting diagram for Cr(III) in this environment. Suppose that the ruby crystal is subjected to high pressure. What do you predict for the variation in the wavelength of absorption of the ruby as a function of pressure? Explain.
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23.85 In 2001, chemists at SUNY-Stony Brook succeeded in synthesizing the complex trans-3Fe(CN)4(CO)242- , which could be a model of complexes that may have played a role in the origin of life. (a) Sketch the structure of the complex. (b) The complex is isolated as a sodium salt. Write the complete name of this salt. (c) What is the oxidation state of Fe in this complex? How many d electrons are associated with the Fe in this complex? (d) Would you expect this complex to be high spin or low spin? Explain. [23.86] When Alfred Werner was developing the field of coordination chemistry, it was argued by some that the optical activity he observed in the chiral complexes he had prepared was because of the presence of carbon atoms in the molecule. To disprove this argument, Werner synthesized a chiral complex of cobalt that had no carbon atoms in it, and he was able to resolve it into its enantiomers. Design a cobalt(III) complex that would be chiral if it could be synthesized and that contains no carbon atoms. (It may not be possible to synthesize the complex you design, but we won’t worry about that for now.) 23.87 Generally speaking, for a given metal and ligand, the stability of a coordination compound is greater for the metal in the +3 rather than in the +2 oxidation state (for metals that
form stable +3 ions in the first place). Suggest an explanation, keeping in mind the Lewis acid–base nature of the metal–ligand bond. 23.88 Many trace metal ions exist in the blood complexed with amino acids or small peptides. The anion of the amino acid glycine (gly), O H2NCH2C
O⫺
can act as a bidentate ligand, coordinating to the metal through nitrogen and oxygen atoms. How many isomers are possible for (a) [Zn(gly)2] (tetrahedral), (b) [Pt(gly)2] (square planar), (c) [Co(gly)3] (octahedral)? Sketch all possible isomers. Use the symbol N O to represent the ligand. [23.89] Suppose that a transition-metal ion was in a lattice in which it was in contact with just two nearby anions, located on opposite sides of the metal. Diagram the splitting of the metal d orbitals that would result from such a crystal field. Assuming a strong field, how many unpaired electrons would you expect for a metal ion with six d electrons? (Hint: Consider the linear axis to be the z-axis)
INTEGRATIVE EXERCISES [23.90] Metallic elements are essential components of many important enzymes operating within our bodies. Carbonic anhydrase, which contains Zn2 + in its active site, is responsible for rapidly interconverting dissolved CO2 and bicarbonate ion, HCO3 - . The zinc in carbonic anhydrase is tetrahedrally coordinated by three neutral nitrogen-containing groups and a water molecule. The coordinated water molecule has a pKa of 7.5, which is crucial for the enzyme’s activity. (a) Draw the active site geometry for the Zn(II) center in carbonic anhydrase, just writing “N” for the three neutral nitrogen ligands from the protein. (b) Compare the pKa of carbonic anhydrase’s active site with that of pure water; which species is more acidic? (c) When the coordinated water to the Zn(II) center in carbonic anhydrase is deprotonated, what ligands are bound to the Zn(II) center? Assume the three nitrogen ligands are unaffected. (d) The pKa of [Zn(H2O)6]2+ is 10. Suggest an explanation for the difference between this pKa and that of carbonic anhydrase. (e) Would you expect carbonic anhydrase to have a deep color, like hemoglobin and other metal-ion containing proteins do? Explain. 23.91 Two different compounds have the formulation CoBr(SO4) # 5 NH3. Compound A is dark violet, and compound B is red-violet. When compound A is treated with AgNO3(aq), no reaction occurs, whereas compound B reacts with AgNO3(aq) to form a white precipitate. When compound A is treated with BaCl2(aq), a white precipitate is formed, whereas compound B has no reaction with BaCl2(aq). (a) Is Co in the same oxidation state in these complexes? (b) Explain the reactivity of compounds A and B with AgNO3(aq) and BaCl2(aq). (c) Are compounds A and B isomers of one another? If so, which category from Figure 23.19 best describes the isomerism observed for these complexes? (d) Would compounds A and B be expected to be strong electrolytes, weak electrolytes, or nonelectrolytes?
23.92 A manganese complex formed from a solution containing potassium bromide and oxalate ion is purified and analyzed. It contains 10.0% Mn, 28.6% potassium, 8.8% carbon, and 29.2% bromine by mass. The remainder of the compound is oxygen. An aqueous solution of the complex has about the same electrical conductivity as an equimolar solution of K4[Fe(CN)6]. Write the formula of the compound, using brackets to denote the manganese and its coordination sphere. 23.93 The E° values for two low-spin iron complexes in acidic solution are as follows: 3Fe(o-phen)343 + (aq) + e - Δ 3Fe(o-phen)342 + (aq) -
3Fe(CN)64 (aq) + e Δ 3Fe(CN)64 (aq) 3-
4-
E° = 1.12 V E° = 0.36 V
(a) Is it thermodynamically favorable to reduce both Fe(III) complexes to their Fe(II) analogs? Explain. (b) Which complex, [Fe(o-phen)3]3+ or [Fe(CN)6]3-, is more difficult to reduce? (c) Suggest an explanation for your answer to (b). 23.94 A palladium complex formed from a solution containing bromide ion and pyridine, C5H5N (a good electron-pair donor), is found on elemental analysis to contain 37.6% bromine, 28.3% carbon, 6.60% nitrogen, and 2.37% hydrogen by mass. The compound is slightly soluble in several organic solvents; its solutions in water or alcohol do not conduct electricity. It is found experimentally to have a zero dipole moment. Write the chemical formula, and indicate its probable structure. 23.95 (a) In early studies it was observed that when the complex [Co(NH3)4Br2]Br was placed in water, the electrical conductivity of a 0.05 M solution changed from an initial value of 191 ohm-1 to a final value of 374 ohm-1 over a period of an hour or so. Suggest an explanation for the observed results. (See Exercise 23.67 for relevant comparison data.) (b) Write a balanced chemical equation to describe the reaction. (c) A 500-mL solution is made up by dissolving 3.87 g of the complex. As soon
Integrative Exercises as the solution is formed, and before any change in conductivity has occurred, a 25.00-mL portion of the solution is titrated with 0.0100 M AgNO3 solution. What volume of AgNO3 solution do you expect to be required to precipitate the free Br -(aq)? (d) Based on the response you gave to part (b), what volume of AgNO3 solution would be required to titrate a fresh 25.00-mL sample of [Co(NH3)4Br2]Br after all conductivity changes have occurred? 23.96 The total concentration of Ca2 + and Mg 2 + in a sample of hard water was determined by titrating a 0.100-L sample of the water with a solution of EDTA4- . The EDTA4- chelates the two cations: Mg 2 + + 3EDTA44- ¡ 3Mg(EDTA)42Ca2 + + 3EDTA44- ¡ 3Ca(EDTA)42It requires 31.5 mL of 0.0104 M 3EDTA44- solution to reach the end point in the titration. A second 0.100-L sample was then treated with sulfate ion to precipitate Ca2 + as calcium sulfate. The Mg 2 + was then titrated with 18.7 mL of 0.0104 M 3EDTA44- . Calculate the concentrations of Mg 2 + and Ca2 + in the hard water in mg>L. 23.97 Carbon monoxide is toxic because it binds more strongly to the iron in hemoglobin (Hb) than does O2, as indicated by these approximate standard free-energy changes in blood: Hb + O2 ¡ HbO2
¢G° = -70 kJ
Hb + CO ¡ HbCO
¢G° = -80 kJ
Using these data, estimate the equilibrium constant at 298 K for the equilibrium HbO2 + CO Δ HbCO + O2 [23.98] The molecule methylamine (CH3NH2) can act as a monodentate ligand. The following are equilibrium reactions and the thermochemical data at 298 K for reactions of methylamine and en with Cd2 + (aq): Cd2 + (aq) + 4 CH3NH2(aq) Δ 3Cd(CH3NH2)442 + (aq) ¢H° = -57.3 kJ;
¢S° = -67.3 J>K;
¢G° = - 37.2 kJ
2+
Cd (aq) + 2 en(aq) Δ 3Cd(en)242 + (aq) ¢H° = -56.5 kJ;
¢S° = +14.1 J>K;
¢G° = - 60.7 kJ
(a) Calculate ¢G° and the equilibrium constant K for the following ligand exchange reaction: 3Cd(CH3NH2)442 + (aq) + 2 en(aq) Δ 3Cd(en)242 + (aq) + 4 CH3NH2(aq)
1003
Based on the value of K in part (a), what would you conclude about this reaction? What concept is demonstrated? (b) Determine the magnitudes of the enthalpic (¢H°) and the entropic (-T¢S°) contributions to ¢G° for the ligand exchange reaction. Explain the relative magnitudes. (c) Based on information in this exercise and in the “A Closer Look” box on the chelate effect, predict the sign of ¢H° for the following hypothetical reaction: 3Cd(CH3NH2)442 + (aq) + 4 NH3(aq) Δ 3Cd(NH3)442 + (aq) + 4 CH3NH2(aq) 23.99 The value of ¢ for the 3CrF643- complex is 182 kJ>mol. Calculate the expected wavelength of the absorption corresponding to promotion of an electron from the lowerenergy to the higher-energy d-orbital set in this complex. Should the complex absorb in the visible range? [23.100] A Cu electrode is immersed in a solution that is 1.00 M in 3Cu(NH3)442 + and 1.00 M in NH3. When the cathode is a standard hydrogen electrode, the emf of the cell is found to be +0.08 V. What is the formation constant for 3Cu(NH3)442 + ? [23.101] The complex 3Ru(EDTA)(H2O)4- undergoes substitution reactions with several ligands, replacing the water molecule with the ligand. In all cases the ruthenium stays in the +3 oxidation state and the ligands use a nitrogen donor atom to bind to the metal. 3Ru(EDTA)(H2O)4- + L ¡ 3Ru(EDTA)L4- + H2O The rate constants for several ligands are as follows: Ligand, L Pyridine SCNCH3CN
k (M-1s-1) 6.3 * 103 2.7 * 102 3.0 * 10
(a) One possible mechanism for this substitution reaction is that the water molecule dissociates from the Ru(III) in the rate-determining step, and then the ligand L binds to Ru(III) in a rapid second step. A second possible mechanism is that L approaches the complex, begins to form a new bond to the Ru(III), and displaces the water molecule, all in a single concerted step. Which of these two mechanisms is more consistent with the data? Explain. (b) What do the results suggest about the relative donor ability of the nitrogens of the three ligands toward Ru(III)? (c) Assuming that the complexes are all low spin, how many unpaired electrons are in each?
WHAT’S AHEAD 24.1 GENERAL CHARACTERISTICS OF ORGANIC MOLECULES
alkynes. Aromatic hydrocarbons have at least one planar ring with delocalized p electrons.
We begin with a review of the structures and reactivities of organic compounds.
24.4 ORGANIC FUNCTIONAL GROUPS
24.2 INTRODUCTION TO HYDROCARBONS We consider hydrocarbons, compounds containing only C and H, including the hydrocarbons called alkanes, which contain only single bonds. We also look at isomers, compounds with identical compositions but different molecular structures.
24.3 ALKENES, ALKYNES, AND AROMATIC
We recognize that a central organizing principle of organic chemistry is the functional group, a group of atoms at which most of the compound’s chemical reactions occur.
24.5 CHIRALITY IN ORGANIC CHEMISTRY We learn that compounds with nonsuperimposable mirror images are chiral and that chirality plays important roles in organic and biological chemistry.
HYDROCARBONS We next explore hydrocarbons with one or more C “ C bonds, called alkenes, and those with one or more C ‚ C bonds, called
24
TO COMMUNICATE WITH OTHER members of their species, insects release chemicals called pheromones into their environment.
24.6 INTRODUCTION TO BIOCHEMISTRY
24.9 LIPIDS
We introduce the chemistry of living organisms, known as biochemistry, biological chemistry, or chemical biology. Important classes of compounds that occur in living systems are proteins, carbohydrates, lipids, and nucleic acids.
We recognize that lipids are a large class of molecules used primarily for energy storage in organisms.
24.7 PROTEINS We learn that proteins are polymers of amino acids linked with amide (also called peptide) bonds. Proteins are used for structural support and as molecular transporters and enzymes.
24.10 NUCLEIC ACIDS We learn that nucleic acids are polymers of nucleotides that contain an organism’s genetic information. Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are polymers composed of nucleotides.
24.8 CARBOHYDRATES We observe that carbohydrates are sugars and polymers of sugars used primarily as fuel by organisms (glucose) or as structural support in plants (cellulose).
THE CHEMISTRY OF LIFE: ORGANIC AND BIOLOGICAL CHEMISTRY INSECTS COMMUNICATE BY RELEASING substances
called pheromones, which they detect with their antennae. There are sex, alarm, defense, and trail pheromones. For example, isoamyl acetate [3-methylbutyl acetate, (CH3)2CHCH2COOCH3] is an alarm pheromone for bees, attracting other bees and provoking them to sting. Mammals, including humans, may also respond to pheromones, although the identity and function of pheromones in humans are not conclusive. Nevertheless, google “pheromone” and you will find hundreds of sources trying to sell you a pheromone, claiming that it will make you irresistible to the opposite sex. 1005
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Although biological systems are almost unimaginably complex, they are nevertheless constructed of molecules of quite modest size, as, for instance, the isoamyl acetate pheromone just described. To understand biology, therefore, we need to understand the chemical behaviors of molecules. This chapter is about the molecules, composed mainly of carbon, hydrogen, oxygen, and nitrogen, that form the basis of organic and biological chemistry. More than 16 million carbon-containing compounds are known. Chemists make thousands of new compounds every year, about 90% of which contain carbon. The study of compounds whose molecules contain carbon constitutes the branch of chemistry known as organic chemistry. This term arose from the eighteenth-century belief that organic compounds could be formed only by living (that is, organic) systems. This idea was disproved in 1828 by the German chemist Friedrich Wöhler when he synthesized urea (H2NCONH2), an organic substance found in the urine of mammals, by heating ammonium cyanate (NH4OCN), an inorganic (“nonliving”) substance. The study of the chemistry of living species is called biological chemistry, chemical biology, or biochemistry. In this chapter, we present some of the elementary aspects of both organic chemistry and biochemistry.
|
24.1 GENERAL CHARACTERISTICS OF ORGANIC MOLECULES What is it about carbon that leads to the tremendous diversity in its compounds and allows it to play such crucial roles in biology and society? Let’s consider some general features of organic molecules and, as we do, review principles we learned in earlier chapters.
The Structures of Organic Molecules Because carbon has four valence electrons ([He]2s22p2), it forms four bonds in virtually all its compounds. When all four bonds are single bonds, the electron pairs are disposed in a tetrahedral arrangement. • (Section 9.2) In the hybridization model, the carbon 2s and 2p orbitals are then sp3 hybridized. • (Section 9.5) When there is one double bond, the arrangement is trigonal planar (sp2 hybridization). With a triple bond, it is linear (sp hybridization). Examples are shown in 쑼 FIGURE 24.1. Almost every organic molecule contains C ¬ H bonds. Because the valence shell of H can hold only two electrons, hydrogen forms only one covalent bond. As a result,
GO FIGURE
What is the geometry around the bottom carbon atom in acetonitrile?
180 109.5 120
씰 FIGURE 24.1 Carbon geometries. The three common geometries around carbon are tetrahedral as in methane (CH4), trigonal planar as in formaldehyde (CH2O), and linear as in acetonitrile (CH3CN). Notice that in all cases each carbon atom forms four bonds.
Tetrahedral 4 single bonds sp3 hybridization
Trigonal planar 2 single bonds 1 double bond sp2 hybridization
Linear 1 single bond 1 triple bond sp hybridization
SECTION 24.1
General Characteristics of Organic Molecules
1007
hydrogen atoms are always located on the surface of organic molecules whereas the C ¬ C bonds form the backbone, or skeleton, of the molecule, as in the propane molecule:
H
H
H
H
C
C
C
H
H
H
H
The Stabilities of Organic Substances Carbon forms strong bonds with a variety of elements, especially H, O, N, and the halogens. • (Section 8.8) Carbon also has an exceptional ability to bond to itself, forming a variety of molecules made up of chains or rings of carbon atoms. Most reactions with low or moderate activation energy (Section 14.5) begin when a region of high electron density on one molecule encounters a region of low electron density on another molecule. The regions of high electron density may be due to the presence of a multiple bond or to the more electronegative atom in a polar bond. Because of their strength and lack of polarity, both C ¬ C single bonds and C ¬ H bonds are relatively unreactive. To better understand the implications of these facts, consider ethanol:
H
H
H
C
C
H
H
O
GO FIGURE
How would replacing OH groups on ascorbic acid with CH3 groups affect the substance’s solubility in (a) polar solvents and (b) nonpolar solvents?
H
The differences in the electronegativity values of C (2.5) and O (3.5) and of O and H (2.1) indicate that the C ¬ O and O ¬ H bonds are quite polar. Thus, many reactions of ethanol involve these bonds while the hydrocarbon portion of the molecule remains intact. A group of atoms such as the C ¬ O ¬ H group, which determines how an organic molecule reacts (in other words, how the molecule functions), is called a functional group. The functional group is the center of reactivity in an organic molecule.
Glucose (C6H12O6)
GIVE IT SOME THOUGHT Which bond is most likely to be the location of a chemical reaction: C “ N, C ¬ C, or C ¬ H?
Solubility and Acid–Base Properties of Organic Substances In most organic substances, the most prevalent bonds are carbon–carbon and carbon–hydrogen, which have low polarity. For this reason, the overall polarity of organic molecules is often low, which makes them generally soluble in nonpolar solvents and not very soluble in water. • (Section 13.3) Organic molecules that are soluble in polar solvents are those that have polar groups on the molecule surface, such as glucose and ascorbic acid (씰 FIGURE 24.2). Organic molecules that have a long, nonpolar part bonded to a polar, ionic part, such as the stearate ion shown in Figure 24.2, function as surfactants and are used in soaps and detergents. • (Section 13.6) The nonpolar part of the molecule extends into a nonpolar medium such as grease or oil, and the polar part extends into a polar medium such as water. Many organic substances contain acidic or basic groups. The most important acidic organic substances are the carboxylic acids, which bear the functional group ¬ COOH. • (Sections 4.3 and 16.10) The most important basic organic substances are amines, which bear the ¬ NH2, ¬ NHR, or ¬ NR2 groups, where R is an organic group made up of carbon and hydrogen atoms. • (Section 16.7) As you read this chapter, you will find many concept links (• ) to related materials in earlier chapters. We strongly encourage you to follow these links and review the earlier material. Doing so will enhance your understanding and appreciation of organic chemistry and biochemistry.
Ascorbic acid (HC6H7O6)
Stearate (C17H35COO) 쑿 FIGURE 24.2 Organic molecules soluble in polar solvents.
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24.2 | INTRODUCTION TO HYDROCARBONS Because carbon compounds are so numerous, it is convenient to organize them into families that have structural similarities. The simplest class of organic compounds is the hydrocarbons, compounds composed of only carbon and hydrogen. The key structural feature of hydrocarbons (and of most other organic substances) is the presence of stable carbon–carbon bonds. Carbon is the only element capable of forming stable, extended chains of atoms bonded through single, double, or triple bonds. Hydrocarbons can be divided into four types, depending on the kinds of carbon–carbon bonds in their molecules. 쑼 TABLE 24.1 shows an example of each type. Alkanes contain only single bonds. Alkenes, also known as olefins, contain at least one C “ C double bond, and alkynes contain at least one C ‚ C triple bond. In aromatic hydrocarbons the carbon atoms are connected in a planar ring structure, joined by both s and delocalized p bonds between carbon atoms. Benzene (C6H6) is the best-known example of an aromatic hydrocarbon. Each type of hydrocarbon exhibits different chemical behaviors, as we will see shortly. The physical properties of all four types, however, are similar in many ways. Because hydrocarbon molecules are relatively nonpolar, they are almost completely insoluble in water but dissolve readily in nonpolar solvents. Their melting points and boiling points are determined by dispersion forces. • (Section 11.2) As a result, hydrocarbons of very low molecular weight, such as C2H6 (bp = -89 °C), are gases at room temperature; those of moderate molecular weight, such as C6H14 (bp = 69 °C), are liquids; and those of high molecular weight, such as C22H46 (mp = 44 °C), are solids. 씰 TABLE 24.2 lists the ten simplest alkanes. Many of these substances are familiar because they are used so widely. Methane is a major component of natural gas and is used for home heating and in gas stoves and water heaters. Propane is the major component of bottled gas used for home heating and cooking in areas where natural gas is not available. Butane is used in disposable lighters and in fuel canisters for gas camping stoves and lanterns. Alkanes with from 5 to 12 carbon atoms per molecule are used to make gasoline. Notice that each succeeding compound in Table 24.2 has an additional CH2 unit.
TABLE 24.1 • The Four Hydrocarbon Types Type
Example
H Alkane
Ethane
109.5
CH3CH3
H C
H
1.54 Å
C
H H
H
H Alkene
Ethylene
CH2
CH2
C
H
Alkyne
Acetylene
CH
C
H
122
H
CH
H
1.34 A
1.21 A
C
H
C
180
H
H C
Aromatic Benzene
C6H6
H
C
120
C H
C C
C
1.39 A
H
H
SECTION 24.2
Introduction to Hydrocarbons
1009
TABLE 24.2 • First Ten Members of the Straight-Chain Alkane Series Molecular Formula
Condensed Structural Formula
Name
Boiling Point (°C)
CH4 C2H6 C3H8 C4H10 C5H12 C6H14 C7H16 C8H18 C9H20 C10H22
CH4 CH3CH3 CH3CH2CH3 CH3CH2CH2CH3 CH3CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH2CH2CH3 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane
-161 -89 -44 -0.5 36 68 98 125 151 174
The formulas for the alkanes given in Table 24.2 are written in a notation called condensed structural formulas. This notation reveals the way in which atoms are bonded to one another but does not require drawing in all the bonds. For example, the structural formula and the condensed structural formulas for butane (C4H10) are
H
H
H
H
H
C
C
C
C
H
H
H
H
H3C H
CH2
CH2
109.5
s orbital of H sp3 orbital of C
CH3
or
CH3CH2CH2CH3
GIVE IT SOME THOUGHT How many C ¬ H and C ¬ C bonds are formed by the middle carbon atom of propane?
Structures of Alkanes
쑿 FIGURE 24.3 Bonds about carbon in methane. This tetrahedral molecular geometry is found around all carbons in alkanes.
According to the VSEPR model, the molecular geometry about each carbon atom in an alkane is tetrahedral. •(Section 9.2) The bonding may be described as involving sp3-hybridized orbitals on the carbon, as pictured in 씰 FIGURE 24.3 for methane. •(Section 9.5) Rotation about a carbon–carbon single bond is relatively easy and occurs rapidly at room temperature. To visualize such rotation, imagine grasping either methyl group of the propane molecule in 씰 FIGURE 24.4 and rotating the group relative to the rest of the molecule. Because motion of this sort occurs rapidly in alkanes, a long-chain alkane molecule is constantly undergoing motions that cause it to change its shape, something like a length of chain that is being shaken.
Structural Isomers The alkanes in Table 24.2 are called straight-chain hydrocarbons because all the carbon atoms are joined in a continuous chain. Alkanes consisting of four or more carbon atoms can also form branched chains, and when they do, they are called branched-chain hydrocarbons. (The branches in organic molecules are often called side chains.) 씰 TABLE 24.3, for example, shows all the straight-chain and branched-chain alkanes containing four and five carbon atoms. Compounds that have the same molecular formula but different bonding arrangements (and hence different structures) are called structural isomers. Thus, C4H10 has two structural isomers and C5H12 has three. The structural isomers of a given alkane differ slightly from one another in physical properties, as the melting and boiling points in Table 24.3 indicate. The number of possible structural isomers increases rapidly with the number of carbon atoms in the alkane. There are 18 isomers with the molecular formula C8H18, for example, and 75 with the molecular formula C10H22.
쑿 FIGURE 24.4 Rotation about a C ¬ C bond occurs easily and rapidly in all alkanes.
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TABLE 24.3 • Isomers of C4H10 and C5H12 Systematic Name (Common Name)
Butane (n-butane)
2-Methylpropane (isobutane)
Condensed Structural Formula
Structural Formula
H H C
H H C C
H C H
H
H H
H
H
H
H
H C
C
C H
Space-filling Model
CH3CH2CH2CH3
CH3
H H H C H
CH
CH3
CH3
Melting Point (°C)
Boiling Point (°C)
138 C
0.5 C
159 C
12 C
130 C
36 C
160 C
28 C
16 C
9 C
H
Pentane (n-pentane)
H
H
H H
H
H C
C
C
C H
H
H
H H
C
CH3CH2CH2CH2CH3
H
H 2-Methylbutane (isopentane)
H C H H H H H C
C
C
CH3
C H
CH3
CH
CH2
CH3
H H
H H H H 2,2-Dimethylpropane (neopentane)
H C H
H C H
CH3
H C H C H
C H H
CH3
C
CH3
CH3
H
GIVE IT SOME THOUGHT What evidence can you cite to support the fact that although isomers have the same molecular formula they are in fact different compounds?
Nomenclature of Alkanes In the first column of Table 24.3, the names in parentheses are called the common names. The common name of the isomer with no branches begins with the letter n (indicating the “normal” structure). When one CH3 group branches off the major chain, the common name of the isomer begins with iso-, and when two CH3 groups branch off, the common name begins with neo-. As the number of isomers grows, however, it becomes impossible to find a suitable prefix to denote each isomer by a common name. The need for a systematic means of naming organic compounds was recognized as early as 1892, when an organization called the International Union of Chemistry met in Geneva to formulate rules for naming organic substances. Since that time the task of updating the
SECTION 24.2
Introduction to Hydrocarbons
1011
rules for naming compounds has fallen to the International Union of Pure and Applied Chemistry (IUPAC). Chemists everywhere, regardless of their nationality, subscribe to a common system for naming compounds. The IUPAC names for the isomers of butane and pentane are the ones given first in Table 24.3. These systematic names, as well as those of other organic compounds, have three parts to them: base suffix
prefix
What substituents?
How many carbons?
What family?
The following steps summarize the procedures used to name alkanes, which all have names ending with -ane. We use a similar approach to write the names of other organic compounds. 1. Find the longest continuous chain of carbon atoms, and use the name of this chain (given in Table 24.2) as the base name. Be careful in this step because the longest chain may not be written in a straight line, as in the following structure: 2
1
CHCH3
CH3
CH2 3
CH2
CH2 4
5
CH3 6
2-Methylhexane
Because the longest continuous chain contains six C atoms, this isomer is named as a substituted hexane. Groups attached to the main chain are called substituents because they are substituted in place of an H atom on the main chain. In this molecule the CH3 group not enclosed by the blue outline is the only substituent in the molecule. 2. Number the carbon atoms in the longest chain, beginning with the end nearest a substituent. In our example, we number the C atoms beginning at the upper right because that places the CH3 substituent on C2 of the chain. (If we had numbered from the lower right, the CH3 would be on C5.) The chain is numbered from the end that gives the lower number to the substituent position. 3. Name each substituent. A substituent formed by removing an H atom from an alkane is called an alkyl group. Alkyl groups are named by replacing the -ane ending of the alkane name with -yl. The methyl group (CH3), for example, is derived from methane (CH4) and the ethyl group (C2H5) is derived from ethane (C2H6). 씰 TABLE 24.4 lists six common alkyl groups. 4. Begin the name with the number or numbers of the carbon or carbons to which each substituent is bonded. For our compound, the name 2-methylhexane indicates the presence of a methyl group on C2 of a hexane (six-carbon) chain. 5. When two or more substituents are present, list them in alphabetical order. The presence of two or more of the same substituent is indicated by the prefixes di- (two), tri- (three), tetra- (four), penta- (five), and so forth. The prefixes are ignored in determining the alphabetical order of the substituents: 7 CH
3
CH3
5
CH
4
CH
6 3
CH2 CH
CH3 2 CH 1
CH2CH3
TABLE 24.4 • Condensed Structural Formulas and Common Names for Several Alkyl Groups Group
Name
CH3 ¬ CH3CH2 ¬ CH3CH2CH2 ¬ CH3CH2CH2CH2 ¬ CH3
Methyl Ethyl Propyl Butyl
HC
Isopropyl
CH3 CH3
CH3
CH3
3-Ethyl-2,4,5-trimethylheptane
CH3
C CH3
tert-Butyl
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CHAPTER 24
The Chemistry of Life: Organic and Biological Chemistry SAMPLE EXERCISE 24.1
Naming Alkanes
Give the systematic name for the following alkane: CH3
CH2
CH
CH3
CH3
CH
CH2
CH3
CH2
SOLUTION Analyze We are given the condensed structural formula of an alkane and asked to give its name. Plan Because the hydrocarbon is an alkane, its name ends in -ane. The name of the parent hydrocarbon is based on the longest continuous chain of carbon atoms. Branches are alkyl groups, named after the number of C atoms in the branch and located by counting C atoms along the longest continuous chain. Solve The longest continuous chain of C atoms extends from the upper left CH3 group to the lower left CH3 group and is seven C atoms long: 1
2
CH3
CH2
3
CH3
CH3
4
CH
5
CH3
6
CH
7
CH2 CH2
The parent compound is thus heptane. There are two methyl groups branching off the main chain. Hence, this compound is a dimethylheptane. To specify the location of the two methyl groups, we must number the C atoms from the end that gives the lower two numbers to the carbons bearing side chains. This means that we should start numbering at the upper left carbon. There is a methyl group on C3 and one on C4. The compound is thus 3,4-dimethylheptane. PRACTICE EXERCISE Name the following alkane:
CH3
CH3
CH
CH
CH2
CH3
CH3 Answer: 2,4-dimethylpentane
SAMPLE EXERCISE 24.2
Writing Condensed Structural Formulas
Write the condensed structural formula for 3-ethyl-2-methylpentane. SOLUTION Analyze We are given the systematic name for a hydrocarbon and asked to write its condensed structural formula. Plan Because the name ends in -ane, the compound is an alkane, meaning that all the carbon–carbon bonds are single bonds. The parent hydrocarbon is pentane, indicating five C atoms (Table 24.2). There are two alkyl groups specified, an ethyl group (two carbon atoms, C2H5) and a methyl group (one carbon atom, CH3). Counting from left to right along the fivecarbon chain, the name tells us that the ethyl group is attached to C3 and the methyl group is attached to C2. Solve We begin by writing five C atoms attached by single bonds. These represent the backbone of the parent pentane chain: C¬C¬C¬C¬C We next place a methyl group on the second C and an ethyl group on the third C of the chain. We then add hydrogens to all the other C atoms to make four bonds to each carbon: CH3 CH3
CH
CH
CH2
CH2CH3
CH3
SECTION 24.2
The formula can be written more concisely as CH3CH(CH3)CH(C2H5)CH2CH3 where the branching alkyl groups are indicated in parentheses. PRACTICE EXERCISE Write the condensed structural formula for 2,3-dimethylhexane. CH3 CH3 Answer: CH3CH
CHCH2CH2CH3
or
CH3CH(CH3)CH(CH3)CH2CH2CH3
Cycloalkanes Alkanes that form rings, or cycles, are called cycloalkanes. As 쑼 FIGURE 24.5 illustrates, cycloalkane structures are sometimes drawn as line structures, which are polygons in which each corner represents a CH2 group. This method of representation is similar to that used for benzene rings. •(Section 8.6) (Remember from our benzene discussion that in aromatic structures each vertex represents a CH group, not a CH2 group.) Carbon rings containing fewer than five carbon atoms are strained because the C ¬ C ¬ C bond angles must be less than the 109.5° tetrahedral angle. The amount of strain increases as the rings get smaller. In cyclopropane, which has the shape of an equilateral triangle, the angle is only 60°; this molecule is therefore much more reactive than propane, its straight-chain analog.
Reactions of Alkanes Because they contain only C ¬ C and C ¬ H bonds, most alkanes are relatively unreactive. At room temperature, for example, they do not react with acids, bases, or strong oxidizing agents. Their low chemical reactivity, as noted in Section 24.1, is due primarily to the strength and lack of polarity of C ¬ C and C ¬ H bonds. Alkanes are not completely inert, however. One of their most commercially important reactions is combustion in air, which is the basis of their use as fuels. • (Section 3.2) For example, the complete combustion of ethane proceeds as follows: 2 C2H6(g) + 7 O2(g) ¡ 4 CO2(g) + 6 H2O(l)
¢H° = -2855 kJ
GO FIGURE
The general formula for straight-chain alkanes is CnH2n+2. What is the general formula for cycloalkanes? H2C H2C
CH2 CH2 CH2 CH2
H2C H2C
CH2 CH2 CH2
CH2 H2C
CH2
Cyclohexane
Cyclopentane
Cyclopropane
Each vertex represents one CH2 group
Five vertices five CH2 groups
Three vertices three CH2 groups
쑿 FIGURE 24.5 Condensed structural formulas and line structures for three cycloalkanes.
Introduction to Hydrocarbons
1013
1014
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The Chemistry of Life: Organic and Biological Chemistry
CHEMISTRY PUT TO WORK Gasoline Petroleum, or crude oil, is a mixture of hydrocarbons plus smaller quantities of other organic compounds containing nitrogen, oxygen, or sulfur. The tremendous demand for petroleum to meet the world’s energy needs has led to the tapping of oil wells in such forbidding places as the North Sea and northern Alaska. The usual first step in the refining, or processing, of petroleum is to separate it into fractions on the basis of boiling point (쑼 TABLE 24.5). Because gasoline is the most commercially important of these fractions, various processes are used to maximize its yield. Gasoline is a mixture of volatile alkanes and aromatic hydrocarbons. In a traditional automobile engine, a mixture of air and gasoline vapor is compressed by a piston and then ignited by a spark plug. The burning of the gasoline should create a strong, smooth expansion of gas, forcing the piston outward and imparting force along the driveshaft of the engine. If the gas burns too rapidly, the piston receives a single hard slam rather than a strong, smooth push. The result is a “knocking” or “pinging” sound and a reduction in the efficiency with which energy produced by the combustion is converted to work. The octane number of a gasoline is a measure of its resistance to knocking. Gasolines with high octane numbers burn more smoothly and are thus more effective fuels (씰 FIGURE 24.6). Branched alkanes and aromatic hydrocarbons have higher octane numbers than straight-chain alkanes. The octane number of gasoline is obtained by comparing its knocking characteristics with those of isooctane (2,2,4-trimethylpentane) and heptane. Isooctane is assigned an octane number of 100, and heptane is assigned 0. Gasoline with the same knocking characteristics as a mixture of 91% isooctane and 9% heptane, for instance, is rated as 91 octane.
TABLE 24.5 • Hydrocarbon Fractions from Petroleum Fraction
Size Range of Molecules
Boiling-Point Range (°C) Uses
Gas
C1 to C5
-160 to 30
Gaseous fuel, production of H2 Motor fuel
Straight-run C5 to C12 gasoline Kerosene, C12 to C18 fuel oil
30 to 200 180 to 400
Diesel fuel, furnace fuel, cracking
Lubricants Paraffins
C16 and up C20 and up
Lubricants Candles, matches
Asphalt
C36 and up
350 and up Low-melting solids Gummy residues
Surfacing roads
쑿 FIGURE 24.6 Octane rating. The octane rating of gasoline measures its resistance to knocking when burned in an engine. The octane rating of the gasoline in the foreground is 89.
The gasoline obtained by fractionating petroleum (called straight-run gasoline) contains mainly straight-chain hydrocarbons and has an octane number around 50. To increase its octane rating, it is subjected to a process called reforming, which converts the straightchain alkanes into branched-chain ones. Cracking is used to produce aromatic hydrocarbons and to convert some of the less-volatile fractions of petroleum into compounds suitable for use as automobile fuel. In cracking, the hydrocarbons are mixed with a catalyst and heated to 400 °C to 500 °C. The catalysts used are either clay minerals or synthetic Al2O3–SiO2 mixtures. In addition to forming molecules more suitable for gasoline, cracking results in the formation of such low-molecular-weight hydrocarbons as ethylene and propene. These substances are used in a variety of reactions to form plastics and other chemicals. Adding compounds called either antiknock agents or octane enhancers increases the octane rating of gasoline. Until the mid-1970s the principal antiknock agent was tetraethyl lead, (C2H5)4Pb. It is no longer used, however, because of the environmental hazards of lead and because it poisons catalytic converters. •(Section 14.7 “Chemistry Put to Work: Catalytic Converters”) Aromatic compounds such as toluene (C6H5CH3) and oxygenated hydrocarbons such as ethanol (CH3CH2OH) are now generally used as antiknock agents. RELATED EXERCISES: 24.19 and 24.20
|
24.3 ALKENES, ALKYNES, AND AROMATIC HYDROCARBONS Because alkanes have only single bonds, they contain the largest possible number of hydrogen atoms per carbon atom. As a result, they are called saturated hydrocarbons. Alkenes, alkynes, and aromatic hydrocarbons contain multiple bonds (double, triple, or delocalized p bonds). As a result, they contain less hydrogen than an alkane with the
SECTION 24.3
Alkenes, Alkynes, and Aromatic Hydrocarbons
GO FIGURE
How many isomers are there for propene, C3H6? Red numbers mark longest chain containing C=C
H
CH3 3
CH3
C 2
Methyl groups on same side of double bond
4
3
C
C 1
H
Methylpropene bp 7 C Methyl group branching off longest chain
H
CH3CH2 H
2
CH3 4
C
C 1
CH3 3
H
1-Butene bp 6 C
H
C
1
Methyl groups on opposite sides of double bond
4
C
2
H
cis-2-Butene bp 4 C
H
CH3 3
H
C 2
CH3 1
trans-2-Butene bp 1 C
No methyl group branching off
쑿 FIGURE 24.7 The alkene C4H8 has four structural isomers.
same number of carbon atoms. Collectively, they are called unsaturated hydrocarbons. On the whole, unsaturated molecules are more reactive than saturated ones.
Alkenes Alkenes are unsaturated hydrocarbons that contain at least one C “ C bond. The simplest alkene is CH2 “ CH2, called ethene (IUPAC) or ethylene, which plays important roles as a plant hormone in seed germination and fruit ripening. The next member of the series is CH3 ¬ CH “ CH2, called propene or propylene. Alkenes with four or more carbon atoms have several isomers. For example, the alkene C4H8 has the four structural isomers shown in 쑿 FIGURE 24.7. Notice both their structures and their names. The names of alkenes are based on the longest continuous chain of carbon atoms that contains the double bond. The chain is named by changing the ending of the name of the corresponding alkane from -ane to -ene. The compound on the left in Figure 24.7, for example, has a double bond as part of a three-carbon chain; thus, the parent alkene is propene. The location of the double bond along an alkene chain is indicated by a prefix number that designates the double-bond carbon atom that is nearest an end of the chain. The chain is always numbered from the end that brings us to the double bond sooner and hence gives the smallest-numbered prefix. In propene the only possible location for the double bond is between the first and second carbons; thus, a prefix indicating its location is unnecessary. For butene (Figure 24.7) there are two possible positions for the double bond, either after the first carbon (1-butene) or after the second carbon (2-butene). GIVE IT SOME THOUGHT How many distinct locations are there for a double bond in a five-carbon linear chain?
If a substance contains two or more double bonds, the location of each is indicated by a numerical prefix, and the ending of the name is altered to identify the number of double bonds: diene (two), triene (three), and so forth. For example, CH2 “ CH ¬ CH2 ¬ CH “ CH2 is 1,4-pentadiene. The two isomers on the right in Figure 24.7 differ in the relative locations of their methyl groups. These two compounds are geometric isomers, compounds that have the same molecular formula and the same groups bonded to one another but differ in the spatial arrangement of these groups. • (Section 23.4) In the cis isomer the two methyl groups are on the same side of the double bond, whereas in the trans isomer they are on opposite sides. Geometric isomers possess distinct physical properties and can differ significantly from each other in their chemical behavior.
1015
1016
CHAPTER 24
The Chemistry of Life: Organic and Biological Chemistry Rotation around double bond requires considerable energy to break bond
Overlapping p orbitals create one π bond
H 씰 FIGURE 24.8 Geometric isomers exist because rotation about a carbon–carbon double bond requires too much energy to occur at ordinary temperatures.
H
cis isomer
R
H
R
R
Rotation has destroyed orbital overlap
R
H
R
R
H
H
trans isomer
This end of molecule rotated 90°
Geometric isomerism in alkenes arises because, unlike the C ¬ C bond, the C “ C bond resists twisting. Recall from Section 9.6 that the double bond between two carbon atoms consists of a s and a p bond. 쑿 FIGURE 24.8 shows a cis alkene. The carbon–carbon bond axis and the bonds to the hydrogen atoms and to the alkyl groups (designated R) are all in a plane, and the p orbitals that form the p bond are perpendicular to that plane. As Figure 24.8 shows, rotation around the carbon–carbon double bond requires the p bond to be broken, a process that requires considerable energy (about 250 kJ>mol). Because rotation doesn’t occur easily around the carbon–carbon bond, the cis and trans isomers of an alkene cannot readily interconvert and, therefore, exist as distinct compounds. SAMPLE EXERCISE 24.3
Drawing Isomers
Draw all the structural and geometric isomers of pentene, C5H10, that have an unbranched hydrocarbon chain. SOLUTION Analyze We are asked to draw all the isomers (both structural and geometric) for an alkene with a five-carbon chain. Plan Because the compound is named pentene and not pentadiene or pentatriene, we know that the five-carbon chain contains only one carbon–carbon double bond. Thus, we begin by placing the double bond in various locations along the chain, remembering that the chain can be numbered from either end. After finding the different unique locations for the double bond, we consider whether the molecule can have cis and trans isomers. Solve There can be a double bond after either the first carbon (1-pentene) or second carbon (2-pentene). These are the only two possibilities because the chain can be numbered from either end. Thus, what we might erroneously call 3-pentene is actually 2-pentene, as seen by numbering the carbon chain from the other end: 1
2
3
4
5
C
C
C
C
C
1
2
3
4
5
C
C
C
C
C
1
2
3
4
5
C
C
C
C
C
1
2
3
4
5
C
C
C
C
C
renumbered as
5
4
3
2
1
C
C
C
C
C
renumbered as
5
4
3
2
1
C
C
C
C
C
Because the first C atom in 1-pentene is bonded to two H atoms, there are no cis-trans isomers. There are cis and trans isomers for 2-pentene, however. Thus, the three isomers for pentene are CH2
CH
CH2
CH2
CH3
CH3 H
H
CH3 C H
trans-2-Pentene
C H
cis-2-Pentene
C CH2
CH2 C
1-Pentene
CH3
CH3
SECTION 24.3
Alkenes, Alkynes, and Aromatic Hydrocarbons
(You should convince yourself that cis-3-pentene is identical to cis-2-pentene and trans3-pentene is identical to trans-2-pentene. However, cis-2-pentene and trans-2-pentene are the correct names because they have smaller numbered prefixes.) PRACTICE EXERCISE How many straight-chain isomers are there of hexene, C6H12? Answer: five (1-hexene, cis-2-hexene, trans-2-hexene, cis-3-hexene, trans-3-hexene)
Alkynes Alkynes are unsaturated hydrocarbons containing one or more C ‚ C bonds. The simplest alkyne is acetylene (C2H2), a highly reactive molecule. When acetylene is burned in a stream of oxygen in an oxyacetylene torch, the flame reaches about 3200 K. Because alkynes in general are highly reactive, they are not as widely distributed in nature as alkenes; alkynes, however, are important intermediates in many industrial processes. Alkynes are named by identifying the longest continuous chain containing the triple bond and modifying the ending of the name of the corresponding alkane from -ane to -yne, as shown in Sample Exercise 24.4. SAMPLE EXERCISE 24.4
Naming Unsaturated Hydrocarbons
Name the following compounds: CH3 (a) CH3CH2CH2
CH
CH3 C
(b) CH3CH2CH2CH
C
H
C
CH
CH2CH2CH3 H
SOLUTION Analyze We are given the condensed structural formulas for an alkene and an alkyne and asked to name the compounds. Plan In each case the name is based on the number of carbon atoms in the longest continuous carbon chain that contains the multiple bond. In the alkene, care must be taken to indicate whether cis-trans isomerism is possible and, if so, which isomer is given. Solve (a) The longest continuous chain of carbons that contains the double bond is seven carbons long, so the parent hydrocarbon is heptene. Because the double bond begins at carbon 2 (numbering from the end closer to the double bond), we have 2-heptene. With a methyl group at carbon atom 4, we have 4-methyl-2-heptene. The geometrical configuration at the double bond is cis (that is, the alkyl groups are bonded to the double bond on the same side). Thus, the full name is 4-methyl-cis-2-heptene. (b) The longest continuous chain containing the triple bond has six carbons, so this compound is a derivative of hexyne. The triple bond comes after the first carbon (numbering from the right), making it 1-hexyne. The branch from the hexyne chain contains three carbon atoms, making it a propyl group. Because this substituent is located on C3 of the hexyne chain, the molecule is 3-propyl-1-hexyne. PRACTICE EXERCISE Draw the condensed structural formula for 4-methyl-2-pentyne. Answer: CH3
C
C
CH
CH3
CH3
Addition Reactions of Alkenes and Alkynes The presence of carbon–carbon double or triple bonds in hydrocarbons markedly increases their chemical reactivity. The most characteristic reactions of alkenes and alkynes are addition reactions, in which a reactant is added to the two atoms that form the multiple bond. A simple example is the addition of a halogen to ethylene: H2C
CH2 Br2
H2C
CH2
Br Br
[24.1]
1017
1018
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The Chemistry of Life: Organic and Biological Chemistry
The pair of electrons that forms the p bond in ethylene is uncoupled and is used to form two s bonds to the two bromine atoms. The s bond between the carbon atoms is retained. Addition of H2 to an alkene converts it to an alkane: Ni, 500 °C
CH3CH “ CHCH3 + H2 999: CH3CH2CH2CH3
[24.2]
The reaction between an alkene and H2, referred to as hydrogenation, does not occur readily at ordinary temperatures and pressures. One reason for the lack of reactivity of H2 toward alkenes is the stability of the H2 bond. To promote the reaction, a catalyst is used to assist in rupturing the H ¬ H bond. The most widely used catalysts are finely divided metals on which H2 is adsorbed. • (Section 14.7) Hydrogen halides and water can also add to the double bond of alkenes, as in these reactions of ethylene: CH2 “ CH2 + HBr 99: CH3CH2Br
[24.3]
H2SO4
CH2 “ CH2 + H2O 99: CH3CH2OH
[24.4]
The addition of water is catalyzed by a strong acid, such as H2SO4. The addition reactions of alkynes resemble those of alkenes, as shown in these examples: Cl CH3C
CCH3 Cl2
CH3 C
C Cl
CH3 2-Butyne
[24.5]
trans-2,3-Dichloro-2-butene
Cl Cl CH3C
CCH3 2 Cl2
CH3
C
C
CH3
[24.6]
Cl Cl 2-Butyne SAMPLE EXERCISE 24.5
2,2,3,3-Tetrachlorobutane Identifying the Product of a Hydrogenation Reaction
Write the condensed structural formula for the product of the hydrogenation of 3-methyl1-pentene. SOLUTION Analyze We are asked to predict the compound formed when a particular alkene undergoes hydrogenation (reaction with H2) and to write the condensed structural formula of the product. Plan To determine the condensed structural formula of the product, we must first write the condensed structural formula or Lewis structure of the reactant. In the hydrogenation of the alkene, H2 adds to the double bond, producing an alkane. Solve The name of the starting compound tells us that we have a chain of five C atoms with a double bond at one end (position 1) and a methyl group on C3: CH3 CH2
CH
CH
CH2
CH3
Hydrogenation—the addition of two H atoms to the carbons of the double bond—leads to the following alkane: CH3 CH3
CH2
CH
CH2
CH3
Comment The longest chain in this alkane has five carbon atoms; the product is therefore 3-methylpentane. PRACTICE EXERCISE Addition of HCl to an alkene forms 2-chloropropane. What is the alkene? Answer: propene
SECTION 24.3
1019
Alkenes, Alkynes, and Aromatic Hydrocarbons
A CLOSER LOOK MECHANISM OF ADDITION REACTIONS As the understanding of chemistry has grown, chemists have advanced from simply cataloging reactions known to occur to explaining how they occur. An explanation of how a reaction occurs is called a mechanism. •(Section 14.6) The addition reaction between HBr and an alkene, for instance, is thought to proceed in two steps. In the first step, which is rate determining •(Section 14.6), the HBr attacks the electron-rich double bond, transferring a proton to one of the double-bond carbons. In the reaction of 2-butene with HBr, for example, the first step is
CH3CH
D
CHCH3 HBr
CH3CH
The energy profile for the reaction is shown in 쑼 FIGURE 24.9. The first energy maximum represents the transition state in the first step, and the second maximum represents the transition state in the second step. The energy minimum represents the energies of the +
intermediate species, CH3CH ¬ CH2CH3 and Br - . To show electron movement in reactions like these, chemists often use curved arrows pointing in the direction of electron flow. For the addition of HBr to 2-butene, for example, the shifts in electron positions are shown as
CH3CH H
CHCH3
CH3
H
C
CH2CH3 Br
The electron pair that formed the p bond is used to form the new C ¬ H bond. The second, faster step is addition of Br - to the positively charged carbon. The bromide ion donates a pair of electrons to the carbon, forming the C ¬ Br bond:
CH3CH
D
CH2CH3 Br
CH3CH
[24.8]
Br Because the rate-determining step involves both the alkene and the acid, the rate law for the reaction is second order, first order in the alkene and first order in HBr: Rate = -
¢3CH3CH “ CHCH34 ¢t
C
CH3 Br
fast
CH3
H
H
C
C
CH3
Br H
GO FIGURE
What features of an energy profile allow you to distinguish between an intermediate and a transition state? First transition state Second transition state
CH2CH3
Br D CH3CHCH2CH3
H
slow
[24.7]
= k3CH3CH “ CHCH343HBr4 [24.9]
Energy
Br
H
Br D CH3CH
CHCH3 H
CH3CH CH2CH3 Br CH3CH CHCH3 Intermediates HBr Reactants
Product CH3CHCH2CH3 Br
Reaction pathway 쑿 FIGURE 24.9 Energy profile for addition of HBr to 2-butene. The two maxima tell you that this is a two-step mechanism.
Aromatic Hydrocarbons The simplest aromatic hydrocarbon, benzene (C6H6), is shown in 쑼 FIGURE 24.10 along with some other aromatic hydrocarbons. Benzene is the most important aromatic hydrocarbon, and most of our discussion focuses on it.
CH3
Benzene
Naphthalene
Anthracene
Toluene (Methylbenzene)
Pyrene
씱 FIGURE 24.10 Line formulas and common names of several aromatic compounds. The aromatic rings are represented by hexagons with a circle inscribed inside to denote delocalized p bonds. Each corner represents a carbon atom. Each carbon is bound to three other atoms—either three carbons or two carbons and a hydrogen––so that each carbon has the requisite four bonds.
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CHAPTER 24
The Chemistry of Life: Organic and Biological Chemistry
Stabilization of p Electrons by Delocalization The planar structure of benzene, with its 120° bond angles, suggests a high degree of unsaturation. You might therefore expect benzene to resemble the alkenes and to be highly reactive. Benzene and the other aromatic hydrocarbons, however, are much more stable than alkenes because the p electrons are delocalized in the p orbitals. • (Section 9.6) We can estimate the stabilization of the p electrons in benzene by comparing the energy required to form cyclohexane by adding hydrogen to benzene, to cyclohexene (one double bond) and to 1,4-cyclohexadiene (two double bonds): 3 H2
H 208 kJ/mol
H2
H 120 kJ/mol
2 H2
H 232 kJ/mol
From the second and third reactions, it appears that the energy required to hydrogenate each double bond is roughly 118 kJ> mol for each bond. Benzene contains the equivalent of three double bonds. We might expect, therefore, the energy required to hydrogenate benzene to be about 3 times -118, or -354 kJ>mol, if benzene behaved as though it were “cyclohexatriene,” that is, if it behaved as though it had three isolated double bonds in a ring. Instead, the energy released is 146 kJ less than this, indicating that benzene is more stable than would be expected for three double bonds. The difference of 146 kJ>mol between the “expected” heat (that is, enthalpy) of hydrogenation, -354 kJ>mol, and the observed heat of hydrogenation, -208 kJ/mol, is due to stabilization of the p electrons through delocalization in the p orbitals that extend around the ring.
Substitution Reactions Although aromatic hydrocarbons are unsaturated, they do not readily undergo addition reactions. The delocalized p bonding causes aromatic compounds to behave quite differently from alkenes and alkynes. Benzene, for example, does not add Cl2 or Br2 to its double bonds under ordinary conditions. In contrast, aromatic hydrocarbons undergo substitution reactions relatively easily. In a substitution reaction one hydrogen atom of a molecule is removed and replaced (substituted) by another atom or group of atoms. When benzene is warmed in a mixture of nitric and sulfuric acids, for example, one of the benzene hydrogens is replaced by the nitro group, NO2: NO2 HNO3
H2SO4
Benzene
H2O
[24.10]
Nitrobenzene
More vigorous treatment results in substitution of a second nitro group into the molecule: NO2
NO2 HNO3
H2SO4
H2O NO2
[24.11]
SECTION 24.4
There are three isomers of benzene that contain two nitro groups––ortho-, meta-, and para-dinitrobenzene: NO2 NO2 NO2 NO2 NO2 NO2 ortho-Dinitrobenzene mp 118 C
meta-Dinitrobenzene mp 90 C
para-Dinitrobenzene mp 174 C
In the reaction of Equation 24.11, the principal product is the meta isomer. Bromination of benzene, carried out with FeBr3 as a catalyst, is another substitution reaction: Br Br2 Benzene
FeBr3
HBr
[24.12]
Bromobenzene
In a similar reaction, called the Friedel-Crafts reaction, alkyl groups can be substituted onto an aromatic ring by reacting an alkyl halide with an aromatic compound in the presence of AlCl3 as a catalyst: CH2CH3 CH3CH2Cl Benzene
AlCl3
HCl
[24.13]
Ethylbenzene
GIVE IT SOME THOUGHT When the aromatic hydrocarbon naphthalene, shown in Figure 24.10, reacts with nitric and sulfuric acids, two compounds containing one nitro group are formed. Draw the structures of these two compounds.
24.4 | ORGANIC FUNCTIONAL GROUPS The C “ C double bonds of alkenes and C ‚ C triple bonds of alkynes are just two of many functional groups in organic molecules. As noted earlier, these functional groups each undergo characteristic reactions, and the same is true of all other functional groups. Each kind of functional group often undergoes the same kinds of reactions in every molecule, regardless of the size and complexity of the molecule. Thus, the chemistry of an organic molecule is largely determined by the functional groups it contains. 씰 TABLE 24.6 lists the most common functional groups. Notice that, except for C “ C and C ‚ C, they all contain either O, N, or a halogen atom, X. We can think of organic molecules as being composed of functional groups bonded to one or more alkyl groups. The alkyl groups, which are made of C ¬ C and C ¬ H single bonds, are the less reactive portions of the molecules. In describing general features of organic compounds, chemists often use the designation R to represent any alkyl group: methyl, ethyl, propyl, and so on. Alkanes, for example, which contain no functional group, are represented as R ¬ H. Alcohols, which contain the group ¬ OH, are represented as R ¬ OH. If two or more different alkyl groups are present in a molecule, we designate them R, R¿ , R– , and so forth.
Organic Functional Groups
1021
1022
CHAPTER 24
The Chemistry of Life: Organic and Biological Chemistry
TABLE 24.6 • Common Functional Groups Example Functional Group
Compound Type
Suffix or Prefix
Structural Formula H
C
C
Alkene
-ene
Alkyne
C
-yne
C
H
C
Ethene (Ethylene)
H C
C
H
Ethyne (Acetylene)
O
H
Methanol (Methyl alcohol)
H C
O
H
Alcohol
-ol
H
C
H H C
O
C
Ether
ether
H
Systematic Name (common name)
H
H C
Ball-and-stick Model
C
H O
C
Dimethyl ether
H
H
H H C (X
Haloalkane
X
halo-
H
N
Amine
-amine
H
O C
Chloromethane (Methyl chloride)
Cl
H
halogen)
C
C
Aldehyde
H
-al
H
H
H
C
C
N
H
H
H
H
O
C
C
H
H
O
H
C
C
C
H
Ethylamine
Ethanal (Acetaldehyde)
H O C
C
C
Ketone
-one
H
C
O
H
Carboxylic acid
-oic acid
H
O
C
Ester
H
O
C
C
O
-oate
H
H
O
O N
Amide
Ethanoic acid (Acetic acid)
C
C
H O
-amide
H
H
O
C
C
H
C H
H
C
H
H
O C
Propanone (Acetone)
H
H O
H
N H
H
H
Methyl ethanoate (Methyl acetate)
Ethanamide (Acetamide)
SECTION 24.4
Organic Functional Groups
1023
CH3 CH3
CH
CH3
CH3
OH
C
CH2
CH3
OH
OH
2-Propanol Isopropyl alcohol; rubbing alcohol
CH2 OH 1,2-Ethanediol Ethylene glycol
2-Methyl-2-propanol t-Butyl alcohol
H3C H3C OH
Phenol
CH3
H3C CH2
CH
CH2
OH
OH
OH
H
H
CH3 씱 FIGURE 24.11 Condensed structural formulas of six important alcohols. Common names are given in blue.
HO
1,2,3-Propanetriol Glycerol; glycerin
Cholesterol
Alcohols Alcohols are hydrocarbon derivatives in which one or more hydrogens of a parent hy-
drocarbon have been replaced by the functional group ¬ OH, called either the hydroxyl group or the alcohol group. Note in 쑿 FIGURE 24.11 that the name for an alcohol ends in -ol. The simple alcohols are named by changing the last letter in the name of the corresponding alkane to -ol—for example, ethane becomes ethanol. Where necessary, the location of the OH group is designated by a numeric prefix that indicates the number of the carbon atom bearing the OH group. The O ¬ H bond is polar, so alcohols are much more soluble in polar solvents than are hydrocarbons. The ¬ OH functional group can also participate in hydrogen bonding. As a result, the boiling points of alcohols are much higher than those of their parent alkanes. 씰 FIGURE 24.12 shows several commercial products that consist entirely or in large part of an organic alcohol. The simplest alcohol, methanol (methyl alcohol), has many industrial uses and is produced on a large scale by heating carbon monoxide and hydrogen under pressure in the presence of a metal oxide catalyst: 200 - 300 atm
CO(g) + 2 H2(g) 9999: CH3OH(g) 400 °C
[24.14]
Because methanol has a very high octane rating as an automobile fuel, it is used as a gasoline additive and as a fuel in its own right. Ethanol (ethyl alcohol, C2H5OH) is a product of the fermentation of carbohydrates such as sugars and starches. In the absence of air, yeast cells convert these carbohydrates into ethanol and CO2: yeast
C6H12O6(aq) 99: 2 C2H5OH(aq) + 2 CO2(g)
[24.15]
In the process, the yeast cells derive energy necessary for growth. This reaction is carried out under carefully controlled conditions to produce beer, wine, and other beverages in which ethanol is the active ingredient. The simplest polyhydroxyl alcohol (an alcohol containing more than one OH group) is 1,2-ethanediol (ethylene glycol, HOCH2CH2OH), the major ingredient in automobile antifreeze. Another common polyhydroxyl alcohol is 1,2,3-propanetriol [glycerol, HOCH2CH(OH)CH2OH], a viscous liquid that dissolves readily in water and is used in cosmetics as a skin softener and in foods and candies to keep them moist. Phenol is the simplest compound with an OH group attached to an aromatic ring. One of the most striking effects of the aromatic group is the greatly increased acidity of the OH group. Phenol is about 1 million times more acidic in water than a nonaromatic alcohol. Even so, it is not a very strong acid (Ka = 1.3 * 10 - 10). Phenol is used industrially to make plastics and dyes, and as a topical anesthetic in throat sprays.
쑿 FIGURE 24.12 Everyday alcohols. Many of the products we use every day— from rubbing alcohol to hair spray and antifreeze—are composed either entirely or mainly of alcohols.
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Cholesterol, shown in Figure 24.11, is a biochemically important alcohol. The OH group forms only a small component of this molecule, so cholesterol is only slightly soluble in water (0.26 g per 100 mL of H2O). Cholesterol is a normal component of our bodies; when present in excessive amounts, however, it may precipitate from solution. It precipitates in the gallbladder to form crystalline lumps called gallstones. It may also precipitate against the walls of veins and arteries and thus contribute to high blood pressure and other cardiovascular problems.
Ethers Compounds in which two hydrocarbon groups are bonded to one oxygen are called ethers. Ethers can be formed from two molecules of alcohol by splitting out a molecule of water. The reaction is catalyzed by sulfuric acid, which takes up water to remove it from the system: H2SO4
CH3CH2 ¬ OH + H ¬ OCH2CH3 999: CH3CH2 ¬ O ¬ CH2CH3 + H2O [24.16] A reaction in which water is split out from two substances is called a condensation reaction. • (Sections 12.8 and 22.8) Both diethyl ether and the cyclic ether tetrahydrofuran are common solvents for organic reactions: CH3CH2
O
CH2
CH2CH3
Diethyl ether
CH2
CH2 CH2 O Tetrahydrofuran (THF)
Aldehydes and Ketones Several of the functional groups listed in Table 24.6 contain the carbonyl group, C “ O. This group, together with the atoms attached to its carbon, defines several important functional groups that we consider in this section. In aldehydes the carbonyl group has at least one hydrogen atom attached: O H
C
O H
Methanal Formaldehyde
CH3
C
H
Ethanal Acetaldehyde
In ketones the carbonyl group occurs at the interior of a carbon chain and is therefore flanked by carbon atoms: O CH3
C
O CH3
Propanone Acetone
CH3
C
CH2CH3
2-Butanone Methyl ethyl ketone
Notice that the systematic names of aldehydes contain -al and that ketone names contain -one. Aldehydes and ketones can be prepared by controlled oxidation of alcohols. Complete oxidation results in formation of CO2 and H2O, as in the burning of methanol: CH3OH(g) +
3 2
O2(g) ¡ CO2(g) + 2 H2O(g)
Controlled partial oxidation to form other organic substances, such as aldehydes and ketones, is carried out by using various oxidizing agents, such as air, hydrogen peroxide (H2O2), ozone (O3), and potassium dichromate (K2Cr2O7).
SECTION 24.4
Organic Functional Groups
1025
GIVE IT SOME THOUGHT Write the condensed structural formula for the ketone that would result from partial oxidation of the alcohol CH2
CHOH
CH2 CH2 CH2
Many compounds found in nature contain an aldehyde or ketone functional group. Vanilla and cinnamon flavorings are naturally occurring aldehydes. Two isomers of the ketone carvone impart the characteristic flavors of spearmint leaves and caraway seeds. Ketones are less reactive than aldehydes and are used extensively as solvents. Acetone, the most widely used ketone, is completely miscible with water, yet it dissolves a wide range of organic substances.
Carboxylic Acids and Esters Carboxylic acids contain the carboxyl functional group, often written COOH. • (Section 16.10) These weak acids are widely distributed in nature and are common in consumer products [씰 FIGURE 24.13(a)]. They are also important in the manufacture of polymers used to make fibers, films, and paints. 쑼 FIGURE 24.14 shows the
(a)
formulas of several carboxylic acids. The common names of many carboxylic acids are based on their historical origins. Formic acid, for example, was first prepared by extraction from ants; its name is derived from the Latin word formica, “ant.” Carboxylic acids can be produced by oxidation of alcohols in which the OH group is attached to a CH2 group. Under appropriate conditions, the aldehyde may be isolated as the first product of oxidation, as in the sequence O CH3CH2OH (O)
CH3CH
Ethanol
H2O
[24.17]
Acetaldehyde
O CH3CH
O (O)
Acetaldehyde
[24.18]
CH3COH Acetic acid
where (O) represents any oxidant that can provide oxygen atoms. The air oxidation of ethanol to acetic acid is responsible for causing wines to turn sour, producing vinegar.
(b) 쑿 FIGURE 24.13 Everyday carboxylic acids and esters. (a) Vinegar contains acetic acid; vitamin C is ascorbic acid; citrus fruits and tomatoes contain citric acid; and aspirin is acetylsalicylic acid (which is both an acid and an ester). (b) Many sunburn lotions contain the ester benzocaine; some nail polish removers contain ethyl acetate; vegetable oils are also esters.
GO FIGURE
Which of these substances have both a carboxylic acid functional group and an alcohol functional group?
O CH3
CH
C
O
OH
OH
H
Lactic acid
C
HO
O
O
HO
C
C
C
CH2
C
CH2
OH
Methanoic acid Formic acid
O
OH
OH Citric acid
O C
OH
O
C
O CH3
O Acetylsalicylic acid Aspirin
CH3
C
O OH
Ethanoic acid Acetic acid
C
OH
Phenyl methanoic acid Benzoic acid
씱 FIGURE 24.14 Structural formulas of common carboxylic acids. The monocarboxylic acids are generally referred to by their common names, given in blue type.
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Acetic acid can also be produced by the reaction of methanol with carbon monoxide in the presence of a rhodium catalyst: O CH3OH CO
catalyst
CH3
C
[24.19]
OH
This reaction involves, in effect, the insertion of a carbon monoxide molecule between the CH3 and OH groups. A reaction of this kind is called carbonylation. Carboxylic acids can undergo condensation reactions with alcohols to form esters: O
O CH3
C
OH HO
Acetic acid
CH2CH3
C
CH3
Ethanol
O
CH2CH3 H2O [24.20]
Ethyl acetate
Esters are compounds in which the H atom of a carboxylic acid is replaced by a carbon-
containing group: O C
O
C
Figure 24.13(b) shows some commercial products containing esters. The name of any ester consists of the name of the group contributed by the alcohol followed by the name of the group contributed by the carboxylic acid, with the -ic replaced by -ate. For example, the ester formed from ethyl alcohol, CH3CH2OH, and butyric acid, CH3(CH2)2COOH, is O CH3CH2CH2C
OCH2CH3
Ethyl butyrate
Notice that the chemical formula generally has the group originating from the acid written first, which is opposite of the way the ester is named. Esters generally have very pleasant odors and are largely responsible for the pleasant aromas of fruit. Pentyl acetate (CH3COOCH2CH2CH2CH2CH3), for example, is responsible for the odor of bananas. An ester treated with an acid or a base in aqueous solution is hydrolyzed; that is, the molecule is split into an alcohol and a carboxylic acid or its anion: O CH3CH2
C
O
CH3 Na OH
Methyl propionate
O CH3CH2
C
O Na CH3OH
Sodium propionate
[24.21]
Methanol
The hydrolysis of an ester in the presence of a base is called saponification, a term that comes from the Latin word for soap, sapon. Naturally occurring esters include fats and oils, and in making soap an animal fat or a vegetable oil is boiled with a strong base. The resultant soap consists of a mixture of salts of long-chain carboxylic acids (called fatty acids), which form during the saponification reaction. • (Section 13.6) Soap has been manufactured and used for thousands of years. Directions for making soap from cassia oil were written on a Babylonian clay tablet around 2200 B.C. For a long time, soap was made by heating animal fat with wood ashes, which contain potassium
SECTION 24.4
carbonate (also known as potash) and made the solution basic. •(Section 16.9) The modern commercial process for making soap usually uses sodium hydroxide as the base. Using potassium hydroxide produces soft or liquid soaps. SAMPLE EXERCISE 24.6
Naming Esters and Predicting Hydrolysis Products
In a basic aqueous solution, esters react with hydroxide ion to form the salt of the carboxylic acid and the alcohol from which the ester is constituted. Name each of the following esters, and indicate the products of their reaction with aqueous base. O
O OCH2CH3
C
(a)
(b) CH3CH2CH2
O
C
SOLUTION Analyze We are given two esters and asked to name them and to predict the products formed when they undergo hydrolysis (split into an alcohol and carboxylate ion) in basic solution. Plan Esters are formed by the condensation reaction between an alcohol and a carboxylic acid. To name an ester, we must analyze its structure and determine the identities of the alcohol and acid from which it is formed. We can identify the alcohol by adding an OH to the alkyl group attached to the O atom of the carboxyl (COO) group. We can identify the acid by adding an H to the O atom of the carboxyl group. We have learned that the first part of an ester name indicates the alcohol portion and the second indicates the acid portion. The name conforms to how the ester undergoes hydrolysis in base, reacting with base to form an alcohol and a carboxylate anion. Solve (a) This ester is derived from ethanol (CH3CH2OH) and benzoic acid (C6H5COOH). Its name is therefore ethyl benzoate. The net ionic equation for reaction of ethyl benzoate with hydroxide ion is O OCH2CH3(aq) OH(aq)
C
O O(aq)
C
HOCH2CH3(aq)
The products are benzoate ion and ethanol. (b) This ester is derived from phenol (C6H5OH) and butanoic acid (commonly called butyric acid) (CH3CH2CH2COOH). The residue from the phenol is called the phenyl group. The ester is therefore named phenyl butyrate. The net ionic equation for the reaction of phenyl butyrate with hydroxide ion is O CH3CH2CH2C
O
(aq)
OH(aq) O
CH3CH2CH2C
O(aq)
HO
(aq)
The products are butyrate ion and phenol. PRACTICE EXERCISE Write the condensed structural formula for the ester formed from propyl alcohol and propionic acid. O Answer: CH3CH2C
O
CH2CH2CH3
Organic Functional Groups
1027
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The Chemistry of Life: Organic and Biological Chemistry
Amines and Amides Amines are compounds in which one or more of the hydrogens of ammonia (NH3) are replaced by an alkyl group: NH2
CH3CH2NH2
(CH3)3N
Ethylamine
Trimethylamine
Phenylamine Aniline
As we have seen earlier, they are the most common organic bases. • (Section 16.7) An amine with at least one H bonded to N can undergo a condensation reaction with a carboxylic acid to form an amide, which contains the carbonyl group (C“O) attached to N (Table 24.6): O CH3C
O OH H
N(CH3)2
CH3C
N(CH3)2 H2O
[24.22]
We may consider the amide functional group to be derived from a carboxylic acid with an NRR¿ group replacing the OH of the acid, as in these examples: O CH3C
O NH2
C
Ethanamide Acetamide
NH2
Phenylmethanamide Benzamide
The amide linkage O R
C
N
R
H where R and R¿ are organic groups, is the key functional group in proteins, as we will see in Section 24.7.
24.5 | CHIRALITY IN ORGANIC CHEMISTRY A molecule possessing a nonsuperimposable mirror image is termed chiral (Greek cheir, “hand”). • (Section 23.4) Compounds containing carbon atoms with four different attached groups are inherently chiral. A carbon atom with four different attached groups is called a chiral center. For example, consider 2-bromopentane: Br CH3
GO FIGURE
If you replace Br with CH3, will the compound be chiral? Mirror
쑿 FIGURE 24.15 The two enantiomeric forms of 2-bromopentane. The mirror-image isomers are not superimposable on each other.
C
CH2CH2CH3
H All four groups attached to C2 are different, making that carbon a chiral center. 씱 FIGURE 24.15 illustrates the nonsuperimposable mirror images of this molecule. Imagine moving the molecule shown to the left of the mirror over to the right of the mirror. If you then turn it in every possible way, you will conclude that it cannot be superimposed on the molecule shown to the right of the mirror. Nonsuperimposable mirror images are called either optical isomers or enantiomers. •(Section 23.4) Organic chemists use the labels R and S to distinguish the two forms. We need not go into the rules for deciding on the labels. The two members of an enantiomer pair have identical physical properties and identical chemical properties when they react with nonchiral reagents. Only in a chiral environment do they
SECTION 24.7
behave differently from each other. One interesting property of chiral substances is that their solutions may rotate the plane of polarized light, as explained in Section 23.4. Chirality is common in organic substances. It is not often observed, however, because when a chiral substance is synthesized in a typical reaction, the two enantiomers are formed in precisely the same quantity. The resulting mixture is called a racemic mixture, and it does not rotate the plane of polarized light because the two forms rotate the light to equal extents in opposite directions. • (Section 23.4) Many drugs are chiral substances. When a drug is administered as a racemic mixture, often only one enantiomer has beneficial results. The other is either inert, or nearly so, or may even have a harmful effect. For example, the drug (R)-albuterol (씰 FIGURE 24.16) is a bronchodilator used to relieve the symptoms of asthma. The enantiomer (S)albuterol is not only ineffective as a bronchodilator but also actually counters the effects of (R)-albuterol. As another example, the nonsteroidal analgesic ibuprofen is a chiral molecule usually sold as the racemic mixture. However, a preparation consisting of just the more active enantiomer, (S)-ibuprofen (씰 FIGURE 24.17), relieves pain and reduces inflammation more rapidly than the racemic mixture. For this reason, the chiral version of the drug may in time come to replace the racemic one.
Proteins
OH HOH2C
1029
H N
HO (R)-Albuterol
쑿 FIGURE 24.16 (R) -Albuterol. This compound, which acts as a bronchodilator in patients with asthma, is one member of an enantiomer pair. The other member, (S)-albuterol, does not have the same physiological effect.
GIVE IT SOME THOUGHT What are the requirements on the four groups attached to a carbon atom in order that it be a chiral center?
CH3
24.6 | INTRODUCTION TO BIOCHEMISTRY The functional groups discussed in Section 24.4 generate a vast array of molecules with very specific chemical reactivities. Nowhere is this specificity more apparent than in biochemistry—the chemistry of living organisms. Before we discuss specific biochemical molecules, we can make some general observations. Many biologically important molecules are quite large, because organisms build biomolecules from smaller, simpler substances readily available in the biosphere. The synthesis of large molecules requires energy because most of the reactions are endothermic. The ultimate source of this energy is the Sun. Animals have essentially no capacity for using solar energy directly, however, and so depend on plant photosynthesis to supply the bulk of their energy needs. • (Section 23.3) In addition to requiring large amounts of energy, living organisms are highly organized. In thermodynamic terms, this high degree of organization means that the entropy of living systems is much lower than that of the raw materials from which the systems formed. Thus, living systems must continuously work against the spontaneous tendency toward increased entropy. In the “Chemistry and Life” essays that appear throughout this text, we have introduced you to some important biochemical applications of fundamental chemical ideas. The remainder of this chapter will serve as only a brief introduction to other aspects of biochemistry. Nevertheless, you will see some patterns emerging. Hydrogen bonding (Section 11.2), for example, is critical to the function of many biochemical systems, and the geometry of molecules (Section 9.1) can govern their biological importance and activity. Many of the large molecules in living systems are polymers (Section 12.8) of much smaller molecules. These biopolymers can be classified into three broad categories: proteins, polysaccharides (carbohydrates), and nucleic acids. Lipids are another common class of molecules in living systems, but they are usually large molecules, not biopolymers.
24.7 | PROTEINS Proteins are macromolecules present in all living cells. About 50% of your body’s dry
mass is protein. Some proteins are structural components in animal tissues; they are a key part of skin, nails, cartilage, and muscles. Other proteins catalyze reactions, transport oxygen, serve as hormones to regulate specific body processes, and perform other tasks. Whatever their function, all proteins are chemically similar, being composed of smaller molecules called amino acids.
COOH H C CH3
CH CH3
CH2
쑿 FIGURE 24.17 (S)-Ibuprofen. For relieving pain and reducing inflammation, the ability of this enantiomer far outweighs that of the (R) isomer.
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Amino Acids An amino acid is a molecule containing an amine group, —NH2, and a carboxylic acid group, —COOH. The building blocks of all proteins are a-amino acids, where the a (alpha) indicates that the amino group is located on the carbon atom immediately adjacent to the carboxylic acid group. Thus, there is always one carbon atom between the amino group and the carboxylic acid group. The general formula for an a-amino acid is represented by One of about 20 different groups
R H2N
C
a carbon
R
COOH
or
H3N
H
C
COO
H
The doubly ionized form, called a zwitterion, usually predominates at near-neutral pH values. This form is a result of the transfer of a proton from the carboxylic acid group to the amine group. •(Section 16.10: “Chemistry and Life: The Amphiprotic Behavior of Amino Acids”) Amino acids differ from one another in the nature of their R groups. Twenty-two amino acids have been identified in nature, and 씰 FIGURE 24.18 shows the 20 of these 22 that are found in humans. Our bodies can synthesize 10 of these 20 amino acids in sufficient amounts for our needs. The other 10 must be ingested and are called essential amino acids because they are necessary components of our diet. The a-carbon atom of the amino acids, which is the carbon between the amino and carboxylate groups, has four different groups attached to it. The amino acids are thus chiral (except for glycine, which has two hydrogens attached to the central carbon). For historical reasons, the two enantiomeric forms of amino acids are often distinguished by the labels D (from the Latin dexter, “right”) and L (from the Latin laevus, “left”). Nearly all the chiral amino acids found in living organisms have the L configuration at the chiral center. The principal exceptions to the dominance of L amino acids in nature are the proteins that make up the cell walls of bacteria, which can contain considerable quantities of the D isomers.
Polypeptides and Proteins Amino acids are linked together into proteins by amide groups (Table 24.6): O R
C
N
R
[24.23]
H Each amide group is called a peptide bond when it is formed by amino acids. A peptide bond is formed by a condensation reaction between the carboxyl group of one amino acid and the amino group of another amino acid. Alanine and glycine, for example, form the dipeptide glycylalanine: H
H3N
C
O C
H O H
H Glycine (Gly; G)
H O
N
C
C
H
CH3
O
Alanine (Ala; A)
H3N
H
O
C
C
H
H O N
C
C
H
CH3
O H2O
Glycylalanine (Gly–Ala; GA)
SECTION 24.7
1031
Proteins
GO FIGURE
Which group of amino acids has a net positive charge at pH 7? Nonpolar amino acids
CH2 CH3 CH3 CH3 CH
CH3
H
CH3 CH
H Glycine (Gly; G)
H Alanine (Ala; A)
H Valine (Val; V)
S
CH2
CH2
CH3 CH
CH2
CH3
CH2
H3N C COO H3N C COO H3N C COO H3N C COO H3N C COO H3N C COO H Leucine (Leu; L)
Polar amino acids
H Isoleucine (Ile; I)
Aromatic amino acids
CH2 H 2 C CH2
H2N C COO
H Methionine (Met; M)
H Proline (Pro; P)
OH
H N
OH
SH
CH2
H3N
CH2 COO
C
CH3
H3N
C
HC
COO
H3N
C
OH
CH2
COO
H3N
C
CH2
COO H3N
CH2 COO
C
H3N
C
COO
H
H
H
H
H
H
Serine (Ser; S)
Cysteine (Cys; C)
Threonine (Thr; T)
Phenylalanine (Phe; F)
Tyrosine (Tyr; Y)
Tryptophan (Trp; W)
Basic amino acids
NH2
HN NH CH2
NH3
C NH2
CH2
NH
CH2
CH2
CH2
CH2
CH2
Acidic amino acids and their amide derivatives
H3N C COO H3N C COO H3N C COO
C
O
O C
CH2
O
O
O
CH2
CH2
CH2
NH2 C
NH2 C
CH2
CH2
O
CH2
H3N C COO H3N C COO H3N C COO H3N C COO
H
H
H
H
H
H
H
Histidine (His; H)
Lysine (Lys; K)
Arginine (Arg; R)
Aspartic acid (Asp; D)
Glutamic acid (Glu; E)
Asparagine (Asn; N)
Glutamine (Gln; Q)
쑿 FIGURE 24.18 The 20 amino acids found in the human body. The acids are shown in the zwitterionic form in which they exist in water at near-neutral pH values.
The amino acid that furnishes the carboxyl group for peptide-bond formation is named first, with a -yl ending; then the amino acid furnishing the amino group is named. Using the abbreviations shown in Figure 24.18, glycylalanine can be abbreviated as either GlyAla or GA. In this notation, it is understood that the unreacted amino group is on the left and the unreacted carboxyl group on the right. The artificial sweetener aspartame (씰 FIGURE 24.19) is the methyl ester of the dipeptide formed from the amino acids aspartic acid and phenylalanine.
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The Chemistry of Life: Organic and Biological Chemistry
CHAPTER 24
SAMPLE EXERCISE 24.7
Drawing the Structural Formula of a Tripeptide
Draw the structural formula for alanylglycylserine. SOLUTION Analyze We are given the name of a substance with peptide bonds and asked to write its structural formula. Plan The name of this substance suggests that three amino acids—alanine, glycine, and serine—have been linked together, forming a tripeptide. Note that the ending -yl has been added to each amino acid except for the last one, serine. By convention, the sequence of amino acids in peptides and proteins is written from the nitrogen end to the carbon end: The firstnamed amino acid (alanine, in this case) has a free amino group and the last-named one (serine) has a free carboxyl group. Solve We first combine the carboxyl group of alanine with the amino group of glycine to form a peptide bond and then the carboxyl group of glycine with the amino group of serine to form another peptide bond: Amino group
H3N
Carboxyl group
H
O
C
C
CH3
H2N C
C
CH2 C
N
C
H
CH2
C O
C
H
H
C
C
N
C
H
CH2OH
O
Ser S
Gly G
We can abbreviate this tripeptide as either Ala-Gly-Ser or AGS.
CH3
PRACTICE EXERCISE Name the dipeptide
OH
H3N
O Aspartic acid (Asp)
N
Ala A
H O
H O
H O
H O
O
C
C
HOCH2
Phenylalanine (Phe)
쑿 FIGURE 24.19 Sweet stuff. The artificial sweetener aspartame is the methyl ester of a dipeptide.
H
H
O
N
C
C
H
CH2
O
COOH and give the two ways of writing its abbreviation. Answer: serylaspartic acid; Ser-Asp, SD. Polypeptides are formed when a large number of amino acids are linked together by peptide bonds. Proteins are linear (that is, unbranched) polypeptide molecules with molecular weights ranging from about 6000 to over 50 million amu. Because up to 22 different amino acids are linked together in proteins and because proteins consist of hundreds of amino acids, the number of possible arrangements of amino acids within proteins is virtually limitless.
Protein Structure The sequence of amino acids along a protein chain is called its primary structure and gives the protein its unique identity. A change in even one amino acid can alter the biochemical characteristics of the protein. For example, sickle-cell anemia is a genetic disorder resulting from a single replacement in a protein chain in hemoglobin. The chain that is affected contains 146 amino acids. The substitution of an amino acid with a hydrocarbon side chain for one that has an acidic functional group in the side chain alters the solubility properties of the hemoglobin, and normal blood flow is impeded. • (Section 13.6: “Chemistry and Life: Sickle-Cell Anemia”) Proteins in living organisms are not simply long, flexible chains with totally random shapes. Rather, the chains self-assemble into structures based on the intermolecular forces we learned about in Chapter 11. This self-assembling leads to a protein’s secondary structure, which refers to how segments of the protein chain are oriented in a regular pattern, as seen in 씰 FIGURE 24.20.
SECTION 24.7
Proteins
1033
Primary structure
R N
H
C C
N
O
C R
R
O C
H
C N H
C
N C
O
R
O
R
C
C N H
O
C
C
N C
R
O
H C
N H
R
C O
R group represents side chain Secondary structure
Tertiary structure
a-helix
b-sheet Quaternary structure
씱 FIGURE 24.20 The four levels of structure of proteins.
One of the most important and common secondary structure arrangements is the A-helix. As the a-helix of Figure 24.20 shows, the helix is held in position by hydrogen
bonds between amide H atoms and carbonyl O atoms. The pitch of the helix and its diameter must be such that (1) no bond angles are strained and (2) the N ¬ H and C “ O functional groups on adjacent turns are in proper position for hydrogen bonding. An arrangement of this kind is possible for some amino acids along the chain but not for others. Large protein molecules may contain segments of the chain that have the ahelical arrangement interspersed with sections in which the chain is in a random coil. The other common secondary structure of proteins is the beta (B) sheet. Beta sheets are made of two or more strands of peptides that hydrogen-bond from an amide H in one strand to a carbonyl O in the other strand (Figure 24.20). GIVE IT SOME THOUGHT If you heat a protein to break the intramolecular hydrogen bonds, will you maintain the a-helical or b -sheet structure?
Proteins are not active biologically unless they are in a particular shape in solution. The process by which the protein adopts its biologically active shape is called folding. The shape of a protein in its folded form—determined by all the bends, kinks, and sections of rodlike a-helical, b -sheet, or flexible coil components—is called the tertiary structure.
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The Chemistry of Life: Organic and Biological Chemistry
CHAPTER 24
Aldehyde
O
H
H
1
H
2
C
OH
HO
3
C
H
H
4
C
H
5
H
6
1
C
OH
2
C
O
HO
3
C
H
OH
H
4
C
OH
C
OH
H
5
C
OH
C
OH
H
6
C
OH
C
H
H
H
Glucose
Fructose
Ketone
쑿 FIGURE 24.21 Linear structure of the carbohydrates glucose and fructose.
Figure 23.14 shows the tertiary structure of myoglobin, a protein with a molecular weight of about 18,000 amu and containing one heme group. Some sections of this protein consist of a-helices. Myoglobin is a globular protein, one that folds into a compact, roughly spherical shape. Globular proteins are generally soluble in water and are mobile within cells. They have nonstructural functions, such as combating the invasion of foreign objects, transporting and storing oxygen, and acting as catalysts. The fibrous proteins form a second class of proteins. In these substances the long coils align more or less in parallel to form long, water-insoluble fibers. Fibrous proteins provide structural integrity and strength to many kinds of tissue and are the main components of muscle, tendons, and hair. The largest known proteins, in excess of 27,000 amino acids long, are muscle proteins. The tertiary structure of a protein is maintained by many different interactions. Certain foldings of the protein chain lead to lower-energy (more stable) arrangements than do other folding patterns. For example, a globular protein dissolved in aqueous solution folds in such a way that the nonpolar hydrocarbon portions are tucked within the molecule, away from the polar water molecules. Most of the more polar acidic and basic side chains, however, project into the solution, where they can interact with water molecules through ion–dipole, dipole–dipole, or hydrogen-bonding interactions. Some proteins are assemblies of more than one polypeptide chain. Each chain has its own tertiary structure, and two or more of these tertiary subunits aggregate into a larger functional macromolecule. The way the tertiary subunits are arranged is called the quaternary structure of the protein (Figure 24.20). For example, hemoglobin, the oxygen-carrying protein of red blood cells, consists of four tertiary subunits. Each subunit contains a component called a heme with an iron atom that binds oxygen as depicted in Figure 23.15. The quaternary structure is maintained by the same types of interactions that maintain the tertiary structure.
24.8 | CARBOHYDRATES Carbohydrates are an important class of naturally occurring substances found in both plant and animal matter. The name carbohydrate (“hydrate of carbon”) comes from the empirical formulas for most substances in this class, which can be written as Cx(H2O)y. For example, glucose, the most abundant carbohydrate, has the molecular formula C6H12O6, or C6(H2O)6. Carbohydrates are not really hydrates of carbon; rather, they are polyhydroxy aldehydes and ketones. Glucose, for example, is a six-carbon aldehyde sugar, whereas fructose, the sugar that occurs widely in fruit, is a six-carbon ketone sugar (씱 FIGURE 24.21). The glucose molecule, having both alcohol and aldehyde functional groups and a reasonably long and flexible backbone, can form a six-member-ring structure, as shown in 쑼 FIGURE 24.22. In fact, in an aqueous solution only a small percentage of the 6
CH2OH
5
H C
4
HO
O
C
H OH C
3
H CH2OH
4
H
C
O
H
C
H OH
H
C
C
C 2 OH
H
OH
HO 씰 FIGURE 24.22 Cyclic glucose has an a form and a b form.
3
a-Glucose
H C
2
O
1
C H
OH
Open form
6 5
H
1
H 4
C
HO
6
CH2OH
5
C
O
OH
H OH
H
C1
C
C2 H
H
OH
3
b-Glucose
SECTION 24.8
glucose molecules are in the open-chain form. Although the ring is often drawn as if it were planar, the molecules are actually nonplanar because of the tetrahedral bond angles around the C and O atoms of the ring. Figure 24.22 shows that the ring structure of glucose can have two relative orientations. In the a form the OH group on C1 and the CH2OH group on C5 point in opposite directions, and in the b form they point in the same direction. Although the difference between the a and b forms might seem small, it has enormous biological consequences, including the vast difference in properties between starch and cellulose. Fructose can cyclize to form either five- or six-member rings. The five-member ring forms when the C5 OH group reacts with the C2 carbonyl group: 6
6
5
5
CH2OH
H 4
C
C
CH2OH
OH
H
2
OH
O C
4
C
1
CH2OH
3
HO
H
C
O
C
H
OH 2
C
OH 3
HO
C
H
1
CH2OH
H
The six-member ring results from the reaction between the C6 OH group and the C2 carbonyl group. SAMPLE EXERCISE 24.8
Identifying Chiral Centers
How many chiral carbon atoms are there in the open-chain form of glucose (Figure 24.21)? SOLUTION Analyze We are given the structure of glucose and asked to determine the number of chiral carbons in the molecule. Plan A chiral carbon has four different groups attached (Section 24.5). We need to identify those carbon atoms in glucose. Solve Carbons 2, 3, 4, and 5 each have four different groups attached to them: O
O
H
1
C
H
2
C
HO
3
C
H
H
4
C
H
5
C
H
6
C
O
H
1
C
H
2
C
HO
3
C
H
OH
H
4
C
OH
H
5
C
H
6
OH
OH
H
C H
O
H
1
C
H
2
C
HO
3
C
H
OH
H
4
C
OH
H
5
C
H
6
OH
OH
C
H
1
C
H
2
C
OH
HO
3
C
H
OH
H
4
C
OH
OH
H
5
C
OH
H
6
C
OH
OH
OH
H
H
Thus, there are four chiral carbon atoms in the glucose molecule. PRACTICE EXERCISE How many chiral carbon atoms are there in the open-chain form of fructose (Figure 24.21)? Answer: three
Disaccharides Both glucose and fructose are examples of monosaccharides, simple sugars that cannot be broken into smaller molecules by hydrolysis with aqueous acids. Two monosaccharide units can be linked together by a condensation reaction to form a disaccharide. The structures of two common disaccharides, sucrose (table sugar) and lactose (milk sugar), are shown in 씰 FIGURE 24.23.
Carbohydrates
1035
1036
CHAPTER 24
The Chemistry of Life: Organic and Biological Chemistry
CH2OH
CH2OH H
C
O
H
HOCH2 O
C
H OH
H
C
C H
C
C
H
OH
HO
O
HO C C CH2OH
C
Glucose unit
H Fructose unit
C
C
OH
O
C
H OH
H
C
C
OH H
H
C
H
C
C
H
H
C
O
H
H
OH
HO Sucrose
OH
C
HO
H
H O
HOCH2 Lactose
Galactose unit
Glucose unit
쑿 FIGURE 24.23 Two disaccharides.
The word sugar makes us think of sweetness. All sugars are sweet, but they differ in the degree of sweetness we perceive when we taste them. Sucrose is about six times sweeter than lactose, slightly sweeter than glucose, but only about half as sweet as fructose. Disaccharides can be reacted with water (hydrolyzed) in the presence of an acid catalyst to form monosaccharides. When sucrose is hydrolyzed, the mixture of glucose and fructose that forms, called invert sugar,* is sweeter to the taste than the original sucrose. The sweet syrup present in canned fruits and candies is largely invert sugar formed from hydrolysis of added sucrose.
Polysaccharides Polysaccharides are made up of many monosaccharide units joined together. The
most important polysaccharides are starch, glycogen, and cellulose, all three of which are formed from repeating glucose units. Starch is not a pure substance. The term refers to a group of polysaccharides found in plants. Starches serve as a major method of food storage in plant seeds and tubers. Corn, potatoes, wheat, and rice all contain substantial amounts of starch. These plant products serve as major sources of needed food energy for humans. Enzymes in the digestive system catalyze the hydrolysis of starch to glucose. Some starch molecules are unbranched chains, whereas others are branched. 쑼 FIGURE 24.24(a) illustrates an unbranched starch structure. Notice, in particular,
(a)
C HO
CH2OH
CH2OH
CH2OH
CH2OH
C
C
C
C
O
C
C
C
OH
O
C
C
C HO
O
C
C
O C
OH
O
C
C
O
C
OH
CH2OH
C
C C
C
C
C
O
OH
CH2OH
O
C
OH
OH
C
C
OH
OH
CH2OH
(b)
C
C
OH
OH
C
O
O
C
C
OH C
OH
C OH
n
OH
O
C
O C
OH
O
C
C OH
OH C
n
C C O
OH
CH2OH
쑿 FIGURE 24.24 Structures of (a) starch and (b) cellulose. *The term invert sugar comes from the fact that rotation of the plane of polarized light by the glucose-fructose mixture is in the opposite direction, or inverted, from that of the sucrose solution.
SECTION 24.9
that the glucose units are in the a form with the bridging oxygen atoms pointing in one direction and the CH2OH groups pointing in the opposite direction Glycogen is a starchlike substance synthesized in the animal body. Glycogen molecules vary in molecular weight from about 5000 to more than 5 million amu. Glycogen acts as a kind of energy bank in the body. It is concentrated in the muscles and liver. In muscles it serves as an immediate source of energy; in the liver it serves as a storage place for glucose and helps to maintain a constant glucose level in the blood. Cellulose [Figure 24.24(b)] forms the major structural unit of plants. Wood is about 50% cellulose; cotton fibers are almost entirely cellulose. Cellulose consists of an unbranched chain of glucose units, with molecular weights averaging more than 500,000 amu. At first glance this structure looks very similar to that of starch. In cellulose, however, the glucose units are in the b form with each bridging oxygen atom pointing in the same direction as the CH2OH group in the ring to its left. Because the individual glucose units have different relationships to one another in starch and cellulose, enzymes that readily hydrolyze starches do not hydrolyze cellulose. Thus, you might eat a pound of cellulose and receive no caloric value from it even though the heat of combustion per unit mass is essentially the same for both cellulose and starch. A pound of starch, in contrast, would represent a substantial caloric intake. The difference is that the starch is hydrolyzed to glucose, which is eventually oxidized with release of energy. However, enzymes in the body do not readily hydrolyze cellulose, so it passes through the digestive system relatively unchanged. Many bacteria contain enzymes, called cellulases, that hydrolyze cellulose. These bacteria are present in the digestive systems of grazing animals, such as cattle, that use cellulose for food. GIVE IT SOME THOUGHT Which type of linkage, a or b, would you expect to join the sugar molecules of glycogen?
24.9 | LIPIDS Lipids are a diverse class of nonpolar biological molecules used by organisms for longterm energy storage (fats, oils) and as elements of biological structures (phospholipids, cell membranes, waxes).
Fats Fats are lipids derived from glyercol and fatty acids. Glycerol is an alcohol with three OH groups. Fatty acids are carboxylic acids (RCOOH) in which R is a hydrocarbon chain, usually 16 to 19 carbon atoms in length. Glycerol and fatty acids undergo condensation reactions to form ester linkages as shown in 씰 FIGURE 24.25. Three fatty acid molecules join to a glycerol. Although the three fatty acids in a fat can be the same, as they are in Figure 24.25, it is also possible that a fat contains three different fatty acids. Lipids with saturated fatty acids are called saturated fats and are commonly solids at room temperature (such as butter and shortening). Unsaturated fats contain one or more double bonds in their carbon–carbon chains. The cis and trans nomenclature we learned for alkenes applies: Trans fats have H atoms on the opposite sides of the C “ C double bond, and cis fats have H atoms on the same sides of the C “ C double bond. Unsaturated fats (such as olive oil and peanut oil) are usually liquid at room temperature and are more often found in plants. For example, the major component (approximately 60 to 80%) of olive oil is oleic acid, cis-CH3(CH2)7CH “ CH(CH2)7COOH. Oleic acid is an example of a monounsaturated fatty acid, meaning it has only one carbon–carbon double bond in the chain. In contrast, polyunsaturated fatty acids have more than one carbon–carbon double bond in the chain. For humans, trans fats are not nutritionally required, which is why some governments are moving to ban them in foods. How, then, do trans fats end up in our food? The process that converts unsaturated fats (such as oils) into saturated fats (such as shortening) is hydrogenation. • (Section 24.3) The by-products of this hydrogenation process include trans fats.
Lipids
1037
1038
CHAPTER 24
The Chemistry of Life: Organic and Biological Chemistry
GO FIGURE
What structural features of a fat molecule cause it to be insoluble in water? Ester linkage H H
C
O O
C
H C H
O H
C
O
C
H C H
O H
C
O
C
H
From glycerol
H C H
H C H H C H H C H
H C H H C H H C H
H C H H C H H C H
H C H H C H H C H
H C H H C H H C H
H C H H C H H C H
H C H H C H H C H
H C H H C H H C H
H C H H C H H C H
H C H H C H H C H
H C H H C H H C H
H C H H C H H C H
H C H H C H H C H
H C
H
H H C
H
H H C
H
H
From fatty acid (palmitic acid)
쑿 FIGURE 24.25 Structure of a fat.
Some of the fatty acids essential for human health must be available in our diets because our metabolism cannot synthesize them. These essential fatty acids are ones that have the carbon–carbon double bonds either three carbons or six carbons away from the ¬ CH3 end of the chain. These are called omega-3 and omega-6 fatty acids, where omega refers to the last carbon in the chain (the carboxylic acid carbon is considered the first, or alpha, one).
Phospholipids Phospholipids are similar in chemical structure to fats but have only two fatty acids attached to a glycerol. The third alcohol group of glycerol is joined to a phosphate group (씰 FIGURE 24.26). The phosphate group can be also attached to a small charged or polar group, such as choline, as shown in the figure. The diversity in phospholipids is based on differences in their fatty acids and in the groups attached to the phosphate group. In water, phospholipids cluster together with their charged polar heads facing the water and their nonpolar tails facing inward. The phospholipids thus form a bilayer that is a key component of cell membranes (씰 FIGURE 24.27).
24.10 | NUCLEIC ACIDS Nucleic acids are a class of biopolymers that are the chemical carriers of an organism’s genetic information. Deoxyribonucleic acids (DNA) are huge molecules whose molecular weights may range from 6 million to 16 million amu. Ribonucleic acids (RNA) are smaller molecules, with molecular weights in the range of 20,000 to 40,000 amu. Whereas DNA is found primarily in the nucleus of the cell, RNA is found mostly outside the nucleus in the cytoplasm, the nonnuclear material enclosed by the cell membrane. DNA stores the genetic information of the cell and specifies which proteins the cell can synthesize. RNA carries the information stored by DNA out of the cell nucleus into the cytoplasm, where the information is used in protein synthesis.
SECTION 24.10
1039
CH2 Hydrophilic head
Nucleic Acids
N(CH2)3
CH2
Choline
O O
P
O
Phosphate
O CH2
CH2
O
O O
C
Glycerol
O
Hydrophobic tails
C
CH2
Fatty acids
쑿 FIGURE 24.26 Structure of a phospholipid.
GO FIGURE
Why do phospholipids form bilayers but not monolayers in water?
Hydrophilic head
Hydrophobic tail
Water
Water
씱 FIGURE 24.27 The cell membrane. Living cells are encased in membranes typically made of phospholipid bilayers. The bilayer structure is stabilized by the favorable interactions of the hydrophobic tails of the phospholipids, which point away from both the water inside the cell and the water outside the cell, while the charged head groups face the two water environments.
1040
CHAPTER 24
The Chemistry of Life: Organic and Biological Chemistry
NH2 N N-containing
N
base unit
N
O
O
P
O
O Phosphate unit
The monomers of nucleic acids, called nucleotides, are formed from a five-carbon sugar, a nitrogen-containing organic base, and a phosphate group. An example is shown in 씱 FIGURE 24.28. The five-carbon sugar in RNA is ribose, and that in DNA is deoxyribose:
N HOCH2 O
CH2 O C H H C OH
C
HC C H
H
H
H
H
C
C
C
C
OH
H
OH OH
H
H
C
C
C OH
Deoxyribose
Deoxyribose differs from ribose only in having one fewer oxygen atom at carbon 2. There are five nitrogen-containing bases in nucleic acids:
쑿 FIGURE 24.28 A nucleotide. Structure of deoxyadenylic acid, the nucleotide formed from phosphoric acid, deoxyribose, and the organic base adenine.
NH2
NH2
O
N
N N
H
OH H
Ribose
Five-carbon sugar unit
H
HOCH2 O
N H
Adenine (A) DNA RNA
HN H2N
N
N H
Guanine (G) DNA RNA
O
N H
Cytosine (C) DNA RNA
O
O CH3
HN
N
N
O
N H Thymine (T) DNA
HN O
N H
Uracil (U) RNA
The first three bases shown here are found in both DNA and RNA. Thymine occurs only in DNA, and uracil occurs only in RNA. In either nucleic acid, each base is attached to a five-carbon sugar through a bond to the nitrogen atom shown in color. The nucleic acids RNA and DNA are polynucleotides formed by condensation reactions between a phosphoric acid OH group on one nucleotide and a sugar OH group on another nucleotide. Thus, the polynucleotide strand has a backbone GO FIGURE consisting of alternating sugar and phosphate groups with the bases extending Is DNA positively charged, negatively off the chain as side groups (씱 FIGURE 24.29). charged, or neutral in aqueous solution The DNA strands wind together in a double helix (씰 FIGURE 24.30). The at pH 7? two strands are held together by attractions between bases (represented by T, A, C, Base O and G). These attractions involve dispersion forces, dipole–dipole forces, and hyO drogen bonds. •(Section 11.2) As shown in 씰 FIGURE 24.31, the structures of thymine and adenine make them perfect partners for hydrogen bonding. Likewise, O cytosine and guanine form ideal hydrogen-bonding partners. We say that thymine Base O P O and adenine are complementary to each other and cytosine and guanine are O O complementary to each other. In the double-helix structure, therefore, each thymine on one strand is opposite an adenine on the other strand, and each cytoO sine is opposite a guanine. The double-helix structure with complementary bases Base O P O on the two strands is the key to understanding how DNA functions. O O The two strands of DNA unwind during cell division, and new complementary strands are constructed on the unraveling strands (씰 FIGURE 24.32). This O process results in two identical double-helix DNA structures, each containing Base O P O one strand from the original structure and one new strand. This replication alO O lows genetic information to be transmitted when cells divide. The structure of DNA is also the key to understanding protein synthesis, the OH means by which viruses infect cells, and many other problems of central impor쑿 FIGURE 24.29 A polynucleotide. Because tance to modern biology. These themes are beyond the scope of this book. If you the sugar in each nucleotide is deoxyribose, this take courses in the life sciences, however, you will learn a good deal about such polynucleotide is of the form found in DNA. matters.
SECTION 24.10
Nucleic Acids
1041
GO FIGURE
Which pair of complementary bases, AT or GC, would you expect to bind more strongly?
T C
G
CH3
A
T
Sugarphosphate backbone
N
H
N
N
Thymine
T
T
A
H N
H
O
N
H
N
N N N
N G
T C Old strand
C
G
A G
C
A
G
A
G
T A C
C
T A T A G
G
A
G
C
G A T
New strand
C
A T
C
A C
G
Sugarphosphate backbone
G T
C
T
T A
SAMPLE INTEGRATIVE EXERCISE
C
A C
A
G
Old strand
G C
T A T A G
Putting Concepts Together
Pyruvic acid,
CH3
G
쑿 FIGURE 24.31 Hydrogen bonding between complementary bases.
T
T
Guanine
C
T
Sugar
NH
H
Cytosine
쑿 FIGURE 24.30 The DNA double helix.
G
O
Sugar G
Sugar
Adenine
T
A
N
O
G
C
A
NH
N
Sugar
G
C
H
N
T
A
O
O
O
C
C
OH
is formed in the body from carbohydrate metabolism. In muscles, it is reduced to lactic acid in the course of exertion. The acid-dissociation constant for pyruvic acid is 3.2 * 10-3. (a) Why does pyruvic acid have a higher acid-dissociation constant than acetic acid? (b) Would you expect pyruvic acid to exist primarily as the neutral acid or as dissociated ions in muscle tissue, assuming a pH of about 7.4 and an acid concentration of 2 * 10-4 M? (c) What would you predict for the solubility properties of pyruvic acid? Explain. (d) What is the hybridization of
G
C
씱 FIGURE 24.32 DNA replication. The original DNA double helix partially unwinds, and new nucleotides line up on each strand in complementary fashion. Hydrogen bonds help align the new nucleotides with the original DNA chain. When the new nucleotides are joined by condensation reactions, two identical double-helix DNA molecules result.
1042
CHAPTER 24
The Chemistry of Life: Organic and Biological Chemistry each carbon atom in pyruvic acid? (e) Assuming H atoms as the reducing agent, write a balanced chemical equation for the reduction of pyruvic acid to lactic acid (Figure 24.14). (Although H atoms do not exist as such in biochemical systems, biochemical reducing agents deliver hydrogen for such reductions.) SOLUTION (a) The acid-dissociation constant for pyruvic acid should be somewhat greater than that of acetic acid because the carbonyl function on the a-carbon atom of pyruvic acid exerts an electron-withdrawing effect on the carboxylic acid group. In the C ¬ O ¬ H bond system the electrons are shifted from H, facilitating loss of the H as a proton. •(Section 16.10) (b) To determine the extent of ionization, we first set up the ionization equilibrium and equilibrium-constant expression. Using HPv as the symbol for the acid, we have HPv Δ H + + Pv Ka =
3H + 43Pv -4 3HPv4
= 3.2 * 10-3
Let 3Pv - 4 = x. Then the concentration of undissociated acid is 2 * 10-4 - x. The concentration of 3H + 4 is fixed at 4.0 * 10-8 (the antilog of the pH value). Substituting, we obtain 3.2 * 10-3 =
34.0 * 10-843x4 32 * 10-4 - x4
Solving for x, we obtain x33.2 * 10-3 + 4.0 * 10-84 = 6.4 * 10-7. The second term in the brackets is negligible compared to the first, so x = 3Pv - 4 = 6.4 * 10-7>3.2 * 10-3 = 2 * 10-4 M This is the initial concentration of acid, which means that essentially all the acid has dissociated. We might have expected this result because the acid is quite dilute and the acid-dissociation constant is fairly high. (c) Pyruvic acid should be quite soluble in water because it has polar functional groups and a small hydrocarbon component. It is miscible with water, ethanol, and diethyl ether. (d) The methyl group carbon has sp3 hybridization. The carbon of the carbonyl group has sp2 hybridization because of the double bond to oxygen. Similarly, the carboxylic acid carbon is sp2 hybridized. (e) The balanced chemical equation for this reaction is O CH3CCOOH 2 (H)
OH CH3CCOOH H
Essentially, the ketonic functional group has been reduced to an alcohol.
S T R AT E G I E S I N C H E M I S T R Y WHAT NOW? If you are reading this box, you have made it to the end of our text. We congratulate you on the tenacity and dedication that you have exhibited to make it this far! As an epilogue, we offer the ultimate study strategy in the form of a question: What do you plan to do with the knowledge of chemistry that you have gained thus far in your studies? Many of you will enroll in additional courses in chemistry as part of your required curriculum. For others, this will be the last formal course in chemistry that you will take. Regardless of the career path you plan to take—whether it is chemistry, one of the biomedical fields, engineering, the liberal arts, or another field—we hope that this text has increased your appreciation of the chemistry in the world around you. If you pay attention, you will be aware of encounters with chemistry on a daily basis, from food and pharmaceutical labels to gasoline pumps, sports equipment to news reports.
We have also tried to give you a sense that chemistry is a dynamic, continuously changing science. Research chemists synthesize new compounds, develop new reactions, uncover chemical properties that were previously unknown, find new applications for known compounds, and refine theories. The understanding of biological systems in terms of the underlying chemistry has become increasingly important as new levels of complexity are uncovered. You may wish to participate in the fascinating venture of chemical research by taking part in an undergraduate research program. Given all the answers that chemists seem to have, you may be surprised at the large number of questions that they still find to ask. Finally, we hope you have enjoyed using this textbook. We certainly enjoyed putting so many of our thoughts about chemistry on paper. We truly believe it to be the central science, one that benefits all who learn about it and from it.
Chapter Summary and Key Terms
1043
CHAPTER SUMMARY AND KEY TERMS INTRODUCTION AND SECTION 24.1
This chapter introduces
organic chemistry, which is the study of carbon compounds (typically compounds containing carbon–carbon bonds), and biochemistry, which is the study of the chemistry of living organisms. We
have encountered many aspects of organic chemistry in earlier chapters. Carbon forms four bonds in its stable compounds. The C ¬ C single bonds and the C ¬ H bonds tend to have low reactivity. Those bonds that have a high electron density (such as multiple bonds or bonds with an atom of high electronegativity) tend to be the sites of reactivity in an organic compound. These sites of reactivity are called functional groups. The simplest types of organic compounds are hydrocarbons, those composed of only carbon and hydrogen. There are four major kinds of hydrocarbons: alkanes, alkenes, alkynes, and aromatic hydrocarbons. Alkanes are composed of only C ¬ H and C ¬ C single bonds. Alkenes contain one or more carbon–carbon double bonds. Alkynes contain one or more carbon–carbon triple bonds. Aromatic hydrocarbons contain cyclic arrangements of carbon atoms bonded through both s and delocalized p bonds. Alkanes are saturated hydrocarbons; the others are unsaturated. Alkanes may form straight-chain, branched-chain, and cyclic arrangements. Isomers are substances that possess the same molecular formula but differ in the arrangements of atoms. In structural isomers the bonding arrangements of the atoms differ. Different isomers are given different systematic names. The naming of hydrocarbons is based on the longest continuous chain of carbon atoms in the structure. The locations of alkyl groups, which branch off the chain, are specified by numbering along the carbon chain. Alkanes with ring structures are called cycloalkanes. Alkanes are relatively unreactive. They do, however, undergo combustion in air, and their chief use is as sources of heat energy produced by combustion. SECTION 24.2
The names of alkenes and alkynes are based on the longest continuous chain of carbon atoms that contains the multiple bond, and the location of the multiple bond is specified by a numerical prefix. Alkenes exhibit not only structural isomerism but geometric (cistrans) isomerism as well. In geometric isomers the bonds are the same, but the molecules have different geometries. Geometric isomerism is possible in alkenes because rotation about the C “ C double bond is restricted. Alkenes and alkynes readily undergo addition reactions to the carbon–carbon multiple bonds. Additions of acids, such as HBr, proceed via a rate-determining step in which a proton is transferred to one of the alkene or alkyne carbon atoms. Addition reactions are difficult to carry out with aromatic hydrocarbons, but substitution reactions are easily accomplished in the presence of catalysts. SECTION 24.3
SECTION 24.4 The chemistry of organic compounds is dominated by the nature of their functional groups. The functional groups we have considered are
O R
O
R
H
Alcohol
C
H
C
Aldehyde
Alkene R (or H)
O C
C
Alkyne
R
C
C
N
Amide
R
N
R (or H)
Amine
O R
C
O O
H
R
C
O
R
Ester
Carboxylic acid
O R
O Ether
R
R
C
R
Ketone
R, R¿ , and R– represent hydrocarbon groups—for example, methyl (CH3) or phenyl (C6H5). Alcohols are hydrocarbon derivatives containing one or more OH groups. Ethers are formed by a condensation reaction of two molecules of alcohol. Several functional groups contain the carbonyl (C “ O) group, including aldehydes, ketones, carboxylic acids, esters, and amides. Aldehydes and ketones can be produced by oxidation of certain alcohols. Further oxidation of the aldehydes produces carboxylic acids. Carboxylic acids can form esters by a condensation reaction with alcohols, or they can form amides by a condensation reaction with amines. Esters undergo hydrolysis (saponification) in the presence of strong bases. SECTION 24.5 Molecules that possess nonsuperimposable mirror images are termed chiral. The two nonsuperimposable forms of a chiral molecule are called enantiomers. In carbon compounds a chiral center is created when all four groups bonded to a central carbon atom are different, as in 2-bromobutane. Many of the molecules occurring in living systems, such as the amino acids, are chiral and exist in nature in only one enantiomeric form. Many drugs of importance in human medicine are chiral, and the enantiomers may produce very different biochemical effects. For this reason, synthesis of only the effective isomers of chiral drugs has become a high priority.
Many of the molecules that are essential for life are large natural polymers that are constructed from smaller molecules called monomers. Three of these biopolymers are considered in this chapter: proteins, polysaccharides (carbohydrates), and nucleic acids. Proteins are polymers of amino acids. They are the major structural materials in animal systems. All naturally occurring proteins are formed from 22 amino acids, although only 20 are common. The amino acids are linked by peptide bonds. A polypeptide is a polymer formed by linking many amino acids by peptide bonds. Amino acids are chiral substances. Usually only one of the enantiomers is found to be biologically active. Protein structure is determined by the sequence of amino acids in the chain (its primary structure), the coiling or stretching of the chain (its secondary structure), and the overall shape of the complete molecule (its tertiary structure). Two important secondary structures are the A-helix and the B sheet. The process by which a protein assumes its biologically active tertiary structure is called folding. Sometimes several proteins aggregate together to form a quaternary structure. SECTIONS 24.6 AND 24.7
Carbohydrates, which are polyhydroxy aldehydes and ketones, are the major structural constituents of plants and are a source of energy in both plants and animals. Glucose is the most common monosaccharide, or simple sugar. Two monosaccharides can be linked together by means of a condensation reaction to form a disaccharide. Polysaccharides are complex carbohydrates
SECTIONS 24.8 AND 24.9
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The Chemistry of Life: Organic and Biological Chemistry
CHAPTER 24
made up of many monosaccharide units joined together. The three most important polysaccharides are starch, which is found in plants; glycogen, which is found in mammals; and cellulose, which is also found in plants. Lipids are compounds derived from glycerol and fatty acids and include fats and phospholipids. Fatty acids can be saturated, unsaturated, cis, or trans depending on their chemical formulas and structures. Nucleic acids are biopolymers that carry the genetic information necessary for cell reproduction; they also control cell
SECTION 24.10
development through control of protein synthesis. The building blocks of these biopolymers are nucleotides. There are two types of nucleic acids, the ribonucleic acids (RNA) and the deoxyribonucleic acids (DNA). These substances consist of a polymeric backbone of alternating phosphate and ribose or deoxyribose sugar groups with organic bases attached to the sugar molecules. The DNA polymer is a doublestranded helix (double helix) held together by hydrogen bonding between matching organic bases situated across from one another on the two strands. The hydrogen bonding between specific base pairs is the key to gene replication and protein synthesis.
KEY SKILLS • Distinguish among alkanes, alkenes, alkynes, and aromatic hydrocarbons. (Section 24.2) • Draw hydrocarbon structures based on their names and name hydrocarbons based on their structures. (Sections 24.2 and 24.3) • Distinguish between addition reactions and substitution reactions. (Section 24.3) • Know the structures of the functional groups: alkene, alkyne, alcohol, carbonyl, ether, aldehyde, ketone, carboxylic acid, amine, amide. (Section 24.4) • Understand what makes a compound chiral and be able to recognize a chiral substance. (Section 24.5) • Recognize the amino acids and understand how they form peptides and proteins via amide bond formation. (Section 24.7) • Understand the differences among the primary, secondary, tertiary, and quaternary structures of proteins. (Section 24.7) • Be able to explain the difference between a-helix and b -sheet peptide and protein structures. (Section 24.7) • Understand the distinction between starch and cellulose structures. (Section 24.8) • Classify molecules as saccharides or lipids based on their structures. (Sections 24.8 and 24.9) • Understand the difference between a saturated and unsaturated fat. (Section 24.9) • Understand the structure of nucleic acids and the role played by complementary bases in DNA replication. (Section 24.10)
EXERCISES VISUALIZING CONCEPTS 24.1 All the structures shown here have the molecular formula C8H18. Which structures are the same molecule? (Hint: One way to answer this question is to determine the chemical name for each structure.) [Section 24.2] CH3
CH3 (a) CH3CCH2CHCH3
CH3
(b) CH3CHCHCH2
CH3 CH3
O CH3C
OH
(c)
(d)
O
CH2
CH3
CH3CH2C (a)
CH3CHCH3 (d) CH3CHCHCH3
CHCH3
CH
CH3
CH2 CH2
CH3
24.2 Which of these molecules is unsaturated? [Section 24.3] CH3CH2CH2CH3 (a)
CHCH3
24.3 Which of these molecules most readily undergoes an addition reaction? [Section 24.3]
CH3
(c) CH3CHCHCH3
CH3CH
CH2
CH2
CH2
CH2 CH2 (b)
OH
(b) O
CH CH2
CH3CHC
CH2
OH
NH2
(c) (d) 24.4 Which of these compounds would you expect to have the highest boiling point? Explain. [Section 24.4] O O CH3CH
CH3CH2OH
(a)
(b)
CH3C (c)
CH
HCOCH3 (d)
Exercises 24.5 Which of these compounds can be a member of an isomer pair? In each case where isomerism is possible, identify the type or types of isomerism. [Sections 24.2, 24.4] O CH3
O
CH3CHCHC
C
O
1045
24.6 From examination of the molecular models i-v, choose the substance that (a) can be hydrolyzed to form a solution containing glucose, (b) is capable of forming a zwitterion, (c) is one of the four bases present in DNA, (d) reacts with an acid to form an ester, (e) is a lipid. [Sections 24.6–24.10]
OH
NH 3 Cl
(a) CH3CH2CH
CHCH3
(b)
CH3CH2CH3
(c)
(d)
(i)
(ii)
(iv)
(iii)
(v)
INTRODUCTION TO ORGANIC COMPOUNDS; HYDROCARBONS (sections 24.1 and 24.2) 24.7 What are the characteristic hybrid orbitals employed by (a) carbon in an alkane, (b) carbon in a double bond in an alkene, (c) carbon in the benzene ring, (d) carbon in a triple bond in an alkyne? 24.8 What are the approximate bond angles (a) about carbon in an alkane, (b) about a doubly bonded carbon atom in an alkene, (c) about a triply bonded carbon atom in an alkyne? 24.9 Predict the ideal values for the bond angles about each carbon atom in the following molecule. Indicate the hybridization of orbitals for each carbon.
CH3CCCH2COOH 24.10 Identify the carbon atom(s) in the structure shown that has (have) each of the following hybridizations: (a) sp3, (b) sp, (c) sp2. N
C
CH2
CH2
CH
CH
24.11 Are carbon monoxide or ammonia considered organic molecules? Why or why not? 24.12 Organic compounds containing C ¬ O and C ¬ Cl bonds are more reactive than simple alkane hydrocarbons. Considering the comparative values of C ¬ H, C ¬ C, C ¬ O, and C ¬ Cl bond enthalpies (Table 8.4), why is this so? 24.13 (a) What is the difference between a straight-chain and branched-chain alkane? (b) What is the difference between an alkane and an alkyl group? 24.14 What structural features help us identify a compound as (a) an alkane, (b) a cycloalkane, (c) an alkene, (d) an alkyne, (e) a saturated hydrocarbon, (f) an aromatic hydrocarbon? 24.15 Give the the name or condensed structural formula, as appropriate:
CHOH C H
O
(a) H
CH3 H
H
H
H
C
C
C
C
C
H
H
H CH3 H
H
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The Chemistry of Life: Organic and Biological Chemistry CH3
(b) CH3CH2CH2CH2CH2CH2CCH2CHCH3
24.17 Give the name or condensed structural formula, as appropriate: (a) CH3CHCH3
CH2 CH3
CHCH2CH2CH2CH3
CH3
CH3
(c) 2-methylheptane (d) 4-ethyl-2,3-dimethyloctane (e) 1,2-dimethylcyclohexane 24.16 Give the name or condensed structural formula, as appropriate: CH3CH2 CH2CH3
(b) (c) (d) (e)
2,2-dimethylpentane 4-ethyl-1,1-dimethylcyclohexane (CH3)2CHCH2CH2C(CH3)3 CH3CH2CH(C2H5)CH2CH2CH2CH3
24.18 Give the name or condensed structural formula, as appropriate: (a) 3-phenylpentane (b) 2,3-dimethylhexane (c) 2-ethyl-2-methylhepane (d) CH3CH2CH(CH3)CH2CH(CH3)2
(a) CH3CCH2CH CH3 CH3
CH3
(e)
(b) CH3CH2CH2CCH3 CH3CHCH2CH3
(c) 2,5,6-trimethylnonane (d) 3-propyl-4,5-methyldecane (e) 1-ethyl-3-methylcyclohexane
CH3
24.19 What is the octane number of a mixture that is 35% heptane and 65% isooctane? 24.20 Describe two ways in which the octane number of a gasoline consisting of alkanes can be increased.
ALKENES, ALKYNES, AND AROMATIC HYDROCARBONS (section 24.3) 24.21 (a) Why are alkanes said to be saturated? (b) Is C4H6 a saturated hydrocarbon? Why or why not?
(d)
Br
24.22 (a) Is the compound CH3CH“CH2 saturated or unsaturated? Explain. (b) What is wrong with the formula CH3CH2CH“CH3? 24.23 Give the molecular formula of a hydrocarbon containing five carbon atoms that is (a) an alkane, (b) a cycloalkane, (c) an alkene, (d) an alkyne. Which are saturated and which are unsaturated hydrocarbons? 24.24 Give the molecular formula of a cyclic alkane, a cyclic alkene, a linear alkyne, and an aromatic hydrocarbon that in each case contains six carbon atoms. Which are saturated and which are unsaturated hydrocarbons? 24.25 Enediynes are a class of compounds that include some antibiotic drugs. Draw the structure of an “enediyne” fragment that contains six carbons in a row. (Hint: di means “two.”) 24.26 Give the general formula for any cyclic alkene, that is, a cyclic hydrocarbon with one double bond. 24.27 Write the condensed structural formulas for as many alkenes and alkynes as you can think of that have the molecular formula C6H10. 24.28 Draw all the possible noncyclic structural isomers of C5H10. Name each compound. 24.29 Name or write the condensed structural formula for the following compounds: (a) trans -2-pentene (b) 2,5-dimethyl-4-octene CH3
(c) CH3CH2
CH2CHCH2CH3 C
H
C H
Br CH2CH3
(e) HC
CCH2CCH3 CH3
24.30 Name or write the condensed structural formula for the following compounds: (a) 4-methyl-2-pentene (b) cis-2,5-dimethyl-3-hexene (c) ortho-dimethylbenzene (d) HC ‚ CCH2CH3 (e) trans-CH3CH“CHCH2CH2CH2CH3 24.31 Why is geometric isomerism possible for alkenes but not for alkanes and alkynes? 24.32 Draw all structural and geometric isomers of butene and name them. 24.33 Indicate whether each of the following molecules is capable of geometrical (cis-trans) isomerism. For those that are, draw the structures: (a) 1,1-dichloro-1-butene, (b) 2,4-dichloro-2butene, (c) 1,4-dichlorobenzene, (d) 4,5-dimethyl-2-pentyne. 24.34 Draw the three distinct geometric isomers of 2,4-hexadiene. 24.35 (a) What is the difference between a substitution reaction and an addition reaction? Which one is commonly observed with alkenes and which one with aromatic hydrocarbons? (b) Using condensed structural formulas, write the balanced equation for the addition reaction of 2-pentene with Br2 and name the
Exercises resulting compound. (c) Write a balanced chemical equation for the substitution reaction of Cl2 with benzene to make para-dichlorobenzene in the presence of FeCl3 as a catalyst. 24.36 Using condensed structural formulas, write a balanced chemical equation for each of the following reactions: (a) hydrogenation of cyclohexene; (b) addition of H2O to trans-2pentene using H2SO4 as a catalyst (two products); (c) reaction of 2-chloropropane with benzene in the presence of AlCl3. 24.37 (a) When cyclopropane is treated with HI, 1-iodopropane is formed. A similar type of reaction does not occur with cyclopentane or cyclohexane. How do you account for the reactivity of cyclopropane? (b) Suggest a method of preparing ethylbenzene, starting with benzene and ethylene as the only organic reagents. 24.38 (a) One test for the presence of an alkene is to add a small amount of bromine, a red-brown liquid, and look for the disappearance of the red-brown color. This test does not work for detecting the presence of an aromatic hydrocarbon. Explain. (b) Write a series of reactions leading to para-bromoethylben-
24.39
24.40
24.41
24.42
1047
zene, beginning with benzene and using other reagents as needed. What isomeric side products might also be formed? The rate law for addition of Br2 to an alkene is first order in Br2 and first order in the alkene. Does this fact prove that the mechanism of addition of Br2 to an alkene proceeds in the same manner as for addition of HBr? Explain. Describe the intermediate that is thought to form in the addition of a hydrogen halide to an alkene, using cyclohexene as the alkene in your description. The molar heat of combustion of gaseous cyclopropane is -2089 kJ>mol; that for gaseous cyclopentane is -3317 kJ>mol. Calculate the heat of combustion per CH2 group in the two cases, and account for the difference. The heat of combustion of decahydronaphthalene (C10H18) is -6286 kJ>mol. The heat of combustion of naphthalene (C10H8) is -5157 kJ>mol. [In both cases CO2(g) and H2O(l) are the products.] Using these data and data in Appendix C, calculate the heat of hydrogenation of naphthalene. Does this value provide any evidence for aromatic character in naphthalene?
FUNCTIONAL GROUPS AND CHIRALITY (sections 24.4 and 24.5) 24.43 Identify the functional groups in each of the following compounds: (a) H3C
CH2
OH
H N
(c) H3C
CH2CH
CH2
O
(b) O
(d)
O
O
O
(f) CH3C
(e) CH3CH2CH2CH2CHO
CCH2COOH
24.44 Identify the functional groups in each of the following compounds: O (a)
CH2CH2CH2CH2CH2CH3
O
H3C
Cl
(b)
CH2CH2CH2CH3
(c)
N H3C
(d) H C H
H H
C H H
C
H H
C H H
H
C
C
H H
C
H H
O
(e) O (f) CH3CH2CH2CH2
24.47 The IUPAC name for a carboxylic acid is based on the name of the hydrocarbon with the same number of carbon atoms. The ending -oic is appended, as in ethanoic acid, which is the IUPAC name for acetic acid. Draw the structure of the following acids: (a) methanoic acid, (b) pentanoic acid, (c) 2-chloro-3-methyldecanoic acid. 24.48 Aldehydes and ketones can be named in a systematic way by counting the number of carbon atoms (including the carbonyl carbon) that they contain. The name of the aldehyde or ketone is based on the hydrocarbon with the same number of carbon atoms. The ending -al for aldehyde or -one for ketone is added as appropriate. Draw the structural formulas for the following aldehydes or ketones: (a) propanal, (b) 2-pentanone, (c) 3-methyl-2-butanone, (d) 2-methylbutanal.
24.50 Draw the condensed structures of the compounds formed from (a) butanoic acid and methanol, (b) benzoic acid and 2-propanol, (c) propanoic acid and dimethylamine. Name the compound in each case.
H H
24.46 (a) Give the empirical formula and structural formula for a cyclic ether containing four carbon atoms in the ring. (b) Write the structural formula for a straight-chain compound that is a structural isomer of your answer to part (a).
24.49 Draw the condensed structure of the compounds formed by condensation reactions between (a) benzoic acid and ethanol, (b) ethanoic acid and methylamine, (c) acetic acid and phenol. Name the compound in each case.
OH O
24.45 Give the structural formula for (a) an aldehyde that is an isomer of acetone, (b) an ether that is an isomer of 1-propanol.
CH2CH2CH2CH3
24.51 Write a balanced chemical equation using condensed structural formulas for the saponification (base hydrolysis) of (a) methyl propionate, (b) phenyl acetate. 24.52 Write a balanced chemical equation using condensed structural formulas for (a) the formation of butyl propionate from the appropriate acid and alcohol, (b) the saponification (base hydrolysis) of methyl benzoate. 24.53 Would you expect pure acetic acid to be a strongly hydrogenbonded substance? How do the melting and boiling points of the substance (16.7 °C and 118 °C) support your answer?
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The Chemistry of Life: Organic and Biological Chemistry
24.54 Acetic anhydride is formed from acetic acid in a condensation reaction that involves the removal of a molecule of water from between two acetic acid molecules. Write the chemical equation for this process, and show the structure of acetic anhydride. 24.55 Write the condensed structural formula for each of the following compounds: (a) 2-pentanol, (b) 1,2-propanediol, (c) ethyl acetate, (d) diphenyl ketone, (e) methyl ethyl ether.
24.56 Write the condensed structural formula for each of the following compounds: (a) 2-ethyl-1-hexanol, (b) methyl phenyl ketone, (c) para-bromobenzoic acid, (d) ethyl butyl ether, (e) N, N-dimethylbenzamide. 24.57 Draw the structure for 2-bromo-2-chloro-3-methylpentane, and indicate any chiral carbons in the molecule. 24.58 Does 3-chloro-3-methylhexane have optical isomers? Why or why not?
PROTEINS (section 24.7) 24.59 (a) What is an a-amino acid? (b) How do amino acids react to form proteins? (c) Draw the bond that links amino acids together in proteins. What is this called? 24.60 What properties of the side chains (R groups) of amino acids are important in affecting the amino acids’ overall biochemical behavior? Give examples to illustrate your reply. 24.61 Draw the two possible dipeptides formed by condensation reactions between leucine and tryptophan. 24.62 Write a chemical equation for the formation of methionyl glycine from the constituent amino acids. 24.63 (a) Draw the condensed structure of the tripeptide Gly-GlyHis. (b) How many different tripeptides can be made from the amino acids glycine and histidine? Give the abbreviations for each of these tripeptides, using the three-letter and one-letter codes for the amino acids. 24.64 (a) What amino acids would be obtained by hydrolysis of the following tripeptide? O
O
O
H2NCHCNHCHCNHCHCOH (CH3)2CH
(b) How many different tripeptides can be made from glycine, serine, and glutamic acid? Give the abbreviation for each of these tripeptides, using the three-letter codes and one-letter codes for the amino acids. 24.65 (a) Describe the primary, secondary, and tertiary structures of proteins. (b) Quaternary structures of proteins arise if two or more smaller polypeptides or proteins associate with each other to make an overall much larger protein structure. The association is due to the same hydrogen bonding, electrostatic, and dispersion forces we have seen before. Hemoglobin, the protein used to transport oxygen molecules in our blood, is an example of a protein that has quaternary structure. Hemoglobin is a tetramer; it is made of four smaller polypeptides, two “alphas” and two “betas.” (These names do not imply anything about the number of alpha-helices or beta sheets in the individual polypeptides.) What kind of experiments would provide sound evidence that hemoglobin exists as a tetramer and not as one enormous polypeptide chain? You may need to look into the chemical literature to discover techniques that chemists and biochemists use to make these decisions. 24.66 What is the difference between the a-helix and b -sheet secondary structures in proteins?
H2COH H2CCH2COH O
CARBOHYDRATES AND LIPIDS (sections 24.8 and 24.9) 24.67 In your own words, define the following terms: (a) carbohydrate, (b) monosaccharide, (c) disaccharide, (d) polysaccharide. 24.68 What is the difference between a-glucose and b -glucose? Show the condensation of two glucose molecules to form a disaccharide with an a linkage; with a b linkage. 24.69 What is the empirical formula of cellulose? What is the unit that forms the basis of the cellulose polymer? What form of linkage joins these monomeric units? 24.70 What is the empirical formula of glycogen? What is the unit that forms the basis of the glycogen polymer? What form of linkage joins these monomeric units? 24.71 The structural formula for the linear form of D-mannose is
(a) How many chiral carbons are present in the molecule? (b) Draw the structure of the six-member-ring form of this sugar. 24.72 The structural formula for the linear form of galactose is O CH H
C
OH
HO
C
H
HO
C
H
H
C
OH
O CH2OH
CH HO
C
H
(a) How many chiral carbons are present in the molecule? (b) Draw the structure of the six-member-ring form of this sugar.
HO
C
H
24.73 Describe the chemical structures of lipids and phospholipids. Why can phospholipids form a bilayer in water?
H
C
OH
H
C
OH
CH2OH
24.74 Using data from Table 8.4 on bond enthalpies, show that the more C¬H bonds a molecule has compared to C ¬ O and O ¬ H bonds, the more energy it can store.
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Additional Exercises
NUCLEIC ACIDS (section 24.10) 24.75 Adenine and guanine are members of a class of molecules known as purines; they have two rings in their structure. Thymine and cytosine, on the other hand, are pyrimidines, and have only one ring in their structure. Predict which have larger dispersion forces in aqueous solution, the purines or the pyrimidines. 24.76 A nucleoside consists of an organic base of the kind shown in Section 24.10, bound to ribose or deoxyribose. Draw the structure for deoxyguanosine, formed from guanine and deoxyribose.
O H
H3C
N
OH
O
P
O
O
N
O
NH2
O
N
N
O
O
P
O
N
O
N
NH2
O N O
O
24.77 Just as the amino acids in a protein are listed in the order from the amine end to the carboxylic acid end (the protein sequence), the bases in nucleic acids are listed in the order 5¿ to 3¿ , where the numbers refer to the position of the carbons in the sugars (shown here for deoxyribose):
P
O
O
N
O
O
O
H
N
N
O
P
O
O
O
N
N
NH2
O
HO
O
5’ 4’
H
H
3’
2’
H
H H
OH
OH
1’
The base is attached to the sugar at the 1¿ carbon. The 5¿ end of a DNA sequence is a phosphate of an OH group, and the 3¿ end of a DNA sequence is the OH group. What is the DNA sequence for the molecule shown here?
24.78 When samples of double-stranded DNA are analyzed, the quantity of adenine present equals that of thymine. Similarly, the quantity of guanine equals that of cytosine. Explain the significance of these observations. 24.79 Imagine a single DNA strand containing a section with the following base sequence: 5¿-GCATTGGC-3¿ . What is the base sequence of the complementary strand? (The two strands of DNA will come together in an antiparallel fashion; that is, 5¿-TAG-3¿ will bind to 3¿-ATC-5¿ .) 24.80 Explain the chemical differences between DNA and RNA.
ADDITIONAL EXERCISES 24.81 Draw the condensed structural formulas for two different molecules with the formula C3H4O. 24.82 How many structural isomers are there for a five-member straight carbon chain with one double bond? For a six-member straight carbon chain with two double bonds? 24.83 Draw the condensed structural formulas for the cis and trans isomers of 2-pentene. Can cyclopentene exhibit cis-trans isomerism? Explain. 24.84 If a molecule is an “ene-one,” what functional groups must it have? 24.85 Write the structural formulas for as many alcohols as you can think of that have empirical formula C3H6O. 24.86 Identify each of the functional groups in these molecules:
H N
O
(c)
O
N H
(Indigo — a blue dye)
CH3 O
NH
(d)
O OH
(a)
(Acetaminophen — aka Tylenol) (Responsible for the odor of cucumbers)
H H2C N
HO
(b)
H
H3CO
24.87 Write a condensed structural formula for each of the following: (a) an acid with the formula C4H8O2, (b) a cyclic ketone with the formula C5H8O, (c) a dihydroxy compound with the formula C3H8O2, (d) a cyclic ester with the formula C5H8O2. 24.88 Although carboxylic acids and alcohols both contain an ¬ OH group, one is acidic in water and the other is not. Explain the difference. [24.89] Indole smells terrible in high concentrations but has a pleasant floral-like odor when highly diluted. Its structure is H
N (Quinine — an antimalarial drug)
N
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The Chemistry of Life: Organic and Biological Chemistry
CHAPTER 24
The molecule is planar, and the nitrogen is a very weak base, with Kb = 2 * 10-12. Explain how this information indicates that the indole molecule is aromatic. 24.90 Locate the chiral carbon atoms, if any, in each molecule: O (a) HOCH2CH2CCH2OH
OH (b) HOCH2CHCCH2OH O
O
CH3
(c) HOCCHCHC2H5
24.91 Which of the following peptides have a net positive charge at pH 7? (a) Gly-Ser-Lys, (b) Pro-Leu-Ile, (c) Phe-Tyr-Asp. 24.92 Glutathione is a tripeptide found in most living cells. Partial hydrolysis yields Cys-Gly and Glu-Cys. What structures are possible for glutathione? 24.93 Monosaccharides can be categorized in terms of the number of carbon atoms (pentoses have five carbons and hexoses have six carbons) and according to whether they contain an aldehyde (aldo- prefix, as in aldopentose) or ketone group (keto- prefix, as in ketopentose). Classify glucose and fructose in this way. 24.94 Can a DNA strand bind to a complementary RNA strand? Explain.
NH2
INTEGRATIVE EXERCISES 24.95 Explain why the boiling point of ethanol (78 °C) is much higher than that of its isomer, dimethyl ether (-25 °C), and why the boiling point of CH2F2 (-52 °C) is far above that of CH4 (-128 °C).
[24.101] The protein ribonuclease A in its native, or most stable, form is folded into a compact globular shape:
24.100 A typical amino acid with one amino group and one carboxylic acid group, such as serine (Figure 24.18), can exist in water in several ionic forms. (a) Suggest the forms of the amino acid at low pH and at high pH. (b) Amino acids generally have two pKa values, one in the range of 2 to 3 and the other in the range of 9 to 10. Serine, for example, has pKa values of 2.19 and 9.21. Using species such as acetic acid and ammonia as models, suggest the origin of the two pKa values. (c) Glutamic acid is an amino acid that has three pKa’s: 2.10, 4.07, and 9.47. Draw the structure of glutamic acid, and assign each pKa to the appropriate part of the molecule.
S S
24.99 The standard free energy of formation of solid glycine is -369 kJ>mol, whereas that of solid glycylglycine is -488 kJ>mol. What is ¢G ° for the condensation of glycine to form glycylglycine?
S S S S
[24.98] An unknown substance is found to contain only carbon and hydrogen. It is a liquid that boils at 49 °C at 1 atm pressure. Upon analysis it is found to contain 85.7% carbon and 14.3% hydrogen by mass. At 100 °C and 735 torr, the vapor of this unknown has a density of 2.21 g>L. When it is dissolved in hexane solution and bromine water is added, no reaction occurs. What is the identity of the unknown compound?
S
24.97 An organic compound is analyzed and found to contain 66.7% carbon, 11.2% hydrogen, and 22.1% oxygen by mass. The compound boils at 79.6 °C. At 100 °C and 0.970 atm, the vapor has a density of 2.28 g>L. The compound has a carbonyl group and cannot be oxidized to a carboxylic acid. Suggest a structure for the compound.
S
[24.96] An unknown organic compound is found on elemental analysis to contain 68.1% carbon, 13.7% hydrogen, and 18.2% oxygen by mass. It is slightly soluble in water. Upon careful oxidation it is converted into a compound that behaves chemically like a ketone and contains 69.7% carbon, 11.7% hydrogen, and 18.6% oxygen by mass. Indicate two or more reasonable structures for the unknown.
Native ribonuclease A
(a) Does the native form have a lower or higher free energy than the denatured form, in which the protein is an extended chain? (b) What is the sign of the entropy change in going from the denatured to the folded form? (c) In the native form, the molecule has four ¬ S ¬ S ¬ bonds that bridge parts of the chain. What effect do you predict these four linkages to have on the free energy and entropy of the native form relative to the free energy and entropy of a hypothetical folded structure that does not have any ¬ S ¬ S ¬ linkages? Explain. (d) A gentle reducing agent converts the four ¬ S ¬ S ¬ linkages in ribonuclease A to eight ¬ S ¬ H bonds. What effect do you predict this conversion to have on the tertiary structure and entropy of the protein? (e) Which amino acid must be present for ¬ SH bonds to exist in ribonuclease A? 24.102 The monoanion of adenosine monophosphate (AMP) is an intermediate in phosphate metabolism: O A
O
P
OH AMP
OH
O
where A = adenosine. If the pKa for this anion is 7.21, what is the ratio of 3AMP ¬ OH - 4 to 3AMP ¬ O2-4 in blood at pH 7.4?
A P P E N D I X
A
MATHEMATICAL OPERATIONS
A.1
EXPONENTIAL NOTATION
The numbers used in chemistry are often either extremely large or extremely small. Such numbers are conveniently expressed in the form N * 10n where N is a number between 1 and 10, and n is the exponent. Some examples of this exponential notation, which is also called scientific notation, follow. 1,200,000 is 1.2 * 106 0.000604 is 6.04 * 10-4
(read “one point two times ten to the sixth power”) (read “six point zero four times ten to the negative fourth power”) A positive exponent, as in the first example, tells us how many times a number must be multiplied by 10 to give the long form of the number: 1.2 * 106 = 1.2 * 10 * 10 * 10 * 10 * 10 * 10 (six tens) = 1,200,000 It is also convenient to think of the positive exponent as the number of places the decimal point must be moved to the left to obtain a number greater than 1 and less than 10. For example, if we begin with 3450 and move the decimal point three places to the left, we end up with 3.45 * 103. In a related fashion, a negative exponent tells us how many times we must divide a number by 10 to give the long form of the number. 6.04 = 0.000604 10 * 10 * 10 * 10 It is convenient to think of the negative exponent as the number of places the decimal point must be moved to the right to obtain a number greater than 1 but less than 10. For example, if we begin with 0.0048 and move the decimal point three places to the right, we end up with 4.8 * 10-3. In the system of exponential notation, with each shift of the decimal point one place to the right, the exponent decreases by 1: 6.04 * 10-4 =
4.8 * 10-3 = 48 * 10-4 Similarly, with each shift of the decimal point one place to the left, the exponent increases by 1: 4.8 * 10-3 = 0.48 * 10-2 Many scientific calculators have a key labeled EXP or EE, which is used to enter numbers in exponential notation. To enter the number 5.8 * 103 on such a calculator, the key sequence is
冷 5 冷冷 # 冷冷 8 冷冷 EXP 冷 (or 冷 EE 冷)冷 3 冷
On some calculators the display will show 5.8, then a space, followed by 03, the exponent. On other calculators, a small 10 is shown with an exponent 3.
1051
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APPENDIX A Mathematical Operations
To enter a negative exponent, use the key labeled +> - . For example, to enter the number 8.6 * 10-5, the key sequence is
冷 8 冷冷 # 冷冷 6 冷冷 EXP 冷冷 +> - 冷冷 5 冷
When entering a number in exponential notation, do not key in the 10 if you use the EXP or EE button. In working with exponents, it is important to recall that 100 = 1. The following rules are useful for carrying exponents through calculations. 1. Addition and Subtraction In order to add or subtract numbers expressed in exponential notation, the powers of 10 must be the same. (5.22 * 104) + (3.21 * 102) = (522 * 102) + (3.21 * 102) = 525 * 102 (3 significant figures) = 5.25 * 104 (6.25 * 10-2) - (5.77 * 10-3) = (6.25 * 10-2) - (0.577 * 10-2) = 5.67 * 10-2 (3 significant figures) When you use a calculator to add or subtract, you need not be concerned with having numbers with the same exponents because the calculator automatically takes care of this matter. 2. Multiplication and Division When numbers expressed in exponential notation are multiplied, the exponents are added; when numbers expressed in exponential notation are divided, the exponent of the denominator is subtracted from the exponent of the numerator. (5.4 * 102)(2.1 * 103) = (5.4)(2.1) * 102 + 3 = 11 * 105 = 1.1 * 106 (1.2 * 105)(3.22 * 10-3) = (1.2)(3.22) * 105 + ( - 3) = 3.9 * 102 3.2 * 105 3.2 = * 105 - 2 = 0.49 * 103 = 4.9 * 102 6.5 6.5 * 102 5.7 * 107 5.7 = * 107 - (-2) = 0.67 * 109 = 6.7 * 108 8.5 8.5 * 10-2 3. Powers and Roots When numbers expressed in exponential notation are raised to a power, the exponents are multiplied by the power. When the roots of numbers expressed in exponential notation are taken, the exponents are divided by the root. (1.2 * 105)3 = (1.2)3 * 105 * 3 = 1.7 * 1015 2 3 2.5 * 106 = 2 3 2.5 * 106>3 = 1.3 * 102 Scientific calculators usually have keys labeled x 2 and 1x for squaring and taking the square root of a number, respectively. To take higher powers or roots, many x calculators have y x and 1 y (or INV y x) keys. For example, to perform the operation x 1 3 7.5 * 10-4 on such a calculator, you would key in 7.5 * 10-4, press the 1 y key (or the INV and then the y x keys), enter the root, 3, and finally press = . The result is 9.1 * 10-2.
APPENDIX A Mathematical Operations
SAMPLE EXERCISE 1
Using Exponential Notation
Perform each of the following operations, using your calculator where possible: (a) Write the number 0.0054 in standard exponential notation. (b) (5.0 * 10-2) + (4.7 * 10-3) (c) (5.98 * 1012)(2.77 * 10-5) 4 1.75 * 10-12 (d) 2 SOLUTION (a) Because we move the decimal point three places to the right to convert 0.0054 to 5.4, the exponent is -3: 5.4 * 10-3 Scientific calculators are generally able to convert numbers to exponential notation using one or two keystrokes; frequently “SCI” for “scientific notation” will convert a number into exponential notation. Consult your instruction manual to see how this operation is accomplished on your calculator. (b) To add these numbers longhand, we must convert them to the same exponent. (5.0 * 10-2) + (0.47 * 10-2) = (5.0 + 0.47) * 10-2 = 5.5 * 10-2 (Note that the result has only two significant figures.) To perform this operation on a calculator, we enter the first number, strike the + key, then enter the second number and strike the = key. (c) Performing this operation longhand, we have (5.98 * 2.77) * 1012 - 5 = 16.6 * 107 = 1.66 * 108 On a scientific calculator, we enter 5.98 * 1012, press the * key, enter 2.77 * 10-5, and press the = key. x (d) To perform this operation on a calculator, we enter the number, press the 1 y key (or the x -3 INV and y keys), enter 4, and press the = key. The result is 1.15 * 10 . PRACTICE EXERCISE Perform the following operations: (a) Write 67,000 in exponential notation, showing two significant figures. (b) (3.378 * 10-3) - (4.97 * 10-5) (c) (1.84 * 1015)(7.45 * 10-2) (d) (6.67 * 10-8)3 Answers: (a) 6.7 * 104, (b) 3.328 * 10-3, (c) 2.47 * 1016, (d) 2.97 * 10-22
A.2
LOGARITHMS
Common Logarithms The common, or base-10, logarithm (abbreviated log) of any number is the power to which 10 must be raised to equal the number. For example, the common logarithm of 1000 (written log 1000) is 3 because raising 10 to the third power gives 1000. 103 = 1000, therefore, log 1000 = 3 Further examples are log 105 = 5 log 1 = 0 Remember that 100 = 1 log 10-2 = -2 In these examples the common logarithm can be obtained by inspection. However, it is not possible to obtain the logarithm of a number such as 31.25 by inspection. The logarithm of 31.25 is the number x that satisfies the following relationship: 10x = 31.25
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APPENDIX A Mathematical Operations
Most electronic calculators have a key labeled LOG that can be used to obtain logarithms. For example, on many calculators we obtain the value of log 31.25 by entering 31.25 and pressing the LOG key. We obtain the following result: log 31.25 = 1.4949 Notice that 31.25 is greater than 10 (101) and less than 100 (102). The value for log 31.25 is accordingly between log 10 and log 100, that is, between 1 and 2.
Significant Figures and Common Logarithms For the common logarithm of a measured quantity, the number of digits after the decimal point equals the number of significant figures in the original number. For example, if 23.5 is a measured quantity (three significant figures), then log 23.5 = 1.371 (three significant figures after the decimal point).
Antilogarithms The process of determining the number that corresponds to a certain logarithm is known as obtaining an antilogarithm. It is the reverse of taking a logarithm. For example, we saw previously that log 23.5 = 1.371. This means that the antilogarithm of 1.371 equals 23.5. log 23.5 = 1.371 antilog 1.371 = 23.5 The process of taking the antilog of a number is the same as raising 10 to a power equal to that number. antilog 1.371 = 101.371 = 23.5 Many calculators have a key labeled 10x that allows you to obtain antilogs directly. On others, it will be necessary to press a key labeled INV (for inverse), followed by the LOG key.
Natural Logarithms Logarithms based on the number e are called natural, or base e, logarithms (abbreviated ln). The natural log of a number is the power to which e (which has the value 2.71828...) must be raised to equal the number. For example, the natural log of 10 equals 2.303. e2.303 = 10, therefore ln 10 = 2.303 Your calculator probably has a key labeled LN that allows you to obtain natural logarithms. For example, to obtain the natural log of 46.8, you enter 46.8 and press the LN key. ln 46.8 = 3.846 The natural antilog of a number is e raised to a power equal to that number. If your calculator can calculate natural logs, it will also be able to calculate natural antilogs. On some calculators there is a key labeled ex that allows you to calculate natural antilogs directly; on others, it will be necessary to first press the INV key followed by the LN key. For example, the natural antilog of 1.679 is given by Natural antilog 1.679 = e1.679 = 5.36 The relation between common and natural logarithms is as follows: ln a = 2.303 log a Notice that the factor relating the two, 2.303, is the natural log of 10, which we calculated earlier.
APPENDIX A Mathematical Operations
Mathematical Operations Using Logarithms Because logarithms are exponents, mathematical operations involving logarithms follow the rules for the use of exponents. For example, the product of z a and z b (where z is any number) is given by
z a # z b = z (a + b) Similarly, the logarithm (either common or natural) of a product equals the sum of the logs of the individual numbers. log ab = log a + log b For the log of a quotient,
ln ab = ln a + ln b
log(a>b) = log a - log b ln(a>b) = ln a - ln b Using the properties of exponents, we can also derive the rules for the logarithm of a number raised to a certain power. log an = n log a log a1>n = (1>n) log a
ln an = n ln a ln a1>n = (1>n) ln a
pH Problems One of the most frequent uses for common logarithms in general chemistry is in working pH problems. The pH is defined as -log3H + 4, where 3H + 4 is the hydrogen ion concentration of a solution. • (Section 16.4) The following sample exercise illustrates this application. SAMPLE EXERCISE 2
Using Logarithms
(a) What is the pH of a solution whose hydrogen ion concentration is 0.015 M? (b) If the pH of a solution is 3.80, what is its hydrogen ion concentration? SOLUTION 1. We are given the value of 3H + 4. We use the LOG key of our calculator to calculate the value of log3H + 4. The pH is obtained by changing the sign of the value obtained. (Be sure to change the sign after taking the logarithm.) 3H + 4 = 0.015 (2 significant figures) log3H + 4 = -1.82 pH = - (-1.82) = 1.82 2. To obtain the hydrogen ion concentration when given the pH, we must take the antilog of -pH. pH = -log3H + 4 = 3.80 log3H + 4 = -3.80 3H + 4 = antilog(-3.80) = 10-3.80 = 1.6 * 10-4 M PRACTICE EXERCISE Perform the following operations: (a) log(2.5 * 10-5), (b) ln 32.7, (c) antilog -3.47, (d) e -1.89. Answers: (a) -4.60, (b) 3.487, (c) 3.4 * 10-4, (d) 1.5 * 10-1
A.3 QUADRATIC EQUATIONS An algebraic equation of the form ax 2 + bx + c = 0 is called a quadratic equation. The two solutions to such an equation are given by the quadratic formula: x =
-b ; 2b2 - 4ac 2a
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APPENDIX A Mathematical Operations
Many calculators today can calculate the solutions to a quadratic equation with one or two keystrokes. Most of the time, x corresponds to the concentration of a chemical species in solution. Only one of the solutions will be a positive number, and that is the one you should use; a “negative concentration” has no physical meaning. SAMPLE EXERCISE 3
Using the Quadratic Formula
Find the values of x that satisfy the equation 2x 2 + 4x = 1. SOLUTION To solve the given equation for x, we must first put it in the form ax 2 + bx + c = 0 and then use the quadratic formula. If 2x 2 + 4x = 1 then 2x 2 + 4x - 1 = 0 Using the quadratic formula, where a = 2, b = 4, and c = -1, we have x =
-4 ; 2(4)(4) - 4(2)( -1) 2(2)
=
-4 ; 216 + 8 -4 ; 224 -4 ; 4.899 = = 4 4 4
x =
-8.899 0.899 = 0.225 and x = = -2.225 4 4
The two solutions are
If this was a problem in which x represented a concentration, we would say x = 0.225 (in the appropriate units), since a negative number for concentration has no physical meaning. TABLE A.1 • Interrelation between Pressure and Temperature Temperature (°C)
Pressure (atm)
20.0 30.0 40.0 50.0
0.120 0.124 0.128 0.132
Pressure (atm)
0.140
0.130
A.4 GRAPHS Often the clearest way to represent the interrelationship between two variables is to graph them. Usually, the variable that is being experimentally varied, called the independent variable, is shown along the horizontal axis (x-axis). The variable that responds to the change in the independent variable, called the dependent variable, is then shown along the vertical axis (y-axis). For example, consider an experiment in which we vary the temperature of an enclosed gas and measure its pressure. The independent variable is temperature, and the dependent variable is pressure. The data shown in TABLE A.1 can be obtained by means of this experiment. These data are shown graphically in FIGURE A.1. The relationship between temperature and pressure is linear. The equation ⌬P for any straight-line graph has the form Slope ⫽ ⌬T ⫽ 0.0123 atm 30.0 ⬚C atm ⫽ 4.10 ⫻ 10⫺4 ⬚C
0.120
Intercept ⫽ 0.112 atm
0.110
0
20.0
40.0 60.0 Temperature (⬚C)
y = mx + b where m is the slope of the line and b is the intercept with the y-axis. In the case of Figure A.1, we could say that the relationship between temperature and pressure takes the form P = mT + b
80.0
FIGURE A.1 A graph of pressure versus temperature yields a straight line for the data.
where P is pressure in atm and T is temperature in °C. As shown in Figure A.1, the slope is 4.10 * 10-4 atm/°C, and the intercept—the point where the line crosses the y-axis—is 0.112 atm. Therefore, the equation for the line is P = a4.10 * 10-4
atm bT + 0.112 atm °C
APPENDIX A Mathematical Operations
A.5
STANDARD DEVIATION
The standard deviation from the mean, s, is a common method for describing precision in experimentally determined data. We define the standard deviation as N
s =
2 a (x i - x)
Q
i=1
N - 1
where N is the number of measurements, x is the average (also called the mean) of the measurements, and xi represents the individual measurements. Electronic calculators with built-in statistical functions can calculate s directly by inputting the individual measurements. A smaller value of s indicates a higher precision, meaning that the data are more closely clustered around the average. The standard deviation has statistical significance. If a large number of measurements is made, 68% of the measured values is expected to be within one standard deviation of the average, assuming only random errors are associated with the measurements. SAMPLE EXERCISE 4
Calculating an Average and Standard Deviation
The percent carbon in a sugar is measured four times: 42.01%, 42.28%, 41.79%, and 42.25%. Calculate (a) the average and (b) the standard deviation for these measurements. SOLUTION (a) The average is found by adding the quantities and dividing by the number of measurements: 42.01 + 42.28 + 41.79 + 42.25 168.33 = = 42.08 4 4 (b) The standard deviation is found using the preceding equation: x =
N
s =
2 a (x i - x)
Q
i=1
N - 1
N
Let’s tabulate the data so the calculation of a (x i - x)2 can be seen clearly. i=1
Percent C
Difference between Measurement and Average, (xi ⴚ x)
42.01 42.28 41.79 42.25
42.01 42.28 41.79 42.25
-
42.08 42.08 42.08 42.08
= = = =
- 0.07 0.20 - 0.29 0.17
Square of Difference, (xi ⴚ x)2 (-0.07)2 = 0.005 (0.20)2 = 0.040 (-0.29)2 = 0.084 (0.17)2 = 0.029
The sum of the quantities in the last column is N
2 a (x i - x) = 0.005 + 0.040 + 0.084 + 0.029 = 0.16
i=1
Thus, the standard deviation is N
s =
2 a (x i - x)
Q
i=1
N - 1
=
0.16 0.16 = = 20.053 = 0.23 A4 - 1 A 3
Based on these measurements, it would be appropriate to represent the measured percent carbon as 42.08 ; 0.23.
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A P P E N D I X
B
PROPERTIES OF WATER
Density:
0.99987 g/mL at 0 °C 1.00000 g/mL at 4 °C 0.99707 g/mL at 25 °C 0.95838 g/mL at 100 °C
Heat (enthalpy) of fusion:
6.008 kJ/mol at 0 °C
Heat (enthalpy) of vaporization:
44.94 kJ/mol at 0 °C 44.02 kJ/mol at 25 °C 40.67 kJ/mol at 100 °C
Ion-product constant, Kw :
1.14 * 10-15 at 0 °C 1.01 * 10-14 at 25 °C 5.47 * 10-14 at 50 °C 2.092 J/g-K = 2.092 J/g # °C for ice at -3 °C 4.184 J/g-K = 4.184 J/g # °C for water at 25 °C 1.841 J/g-K = 1.841 J/g # °C for steam at 100 °C
Specific heat:
Vapor Pressure (torr) at Different Temperatures
1058
T(°C)
P
T(°C)
P
T(°C)
P
T(°C)
P
0 5 10 12 14 16 17 18 19 20
4.58 6.54 9.21 10.52 11.99 13.63 14.53 15.48 16.48 17.54
21 22 23 24 25 26 27 28 29 30
18.65 19.83 21.07 22.38 23.76 25.21 26.74 28.35 30.04 31.82
35 40 45 50 55 60 65 70 80 90
42.2 55.3 71.9 92.5 118.0 149.4 187.5 233.7 355.1 525.8
92 94 96 98 100 102 104 106 108 110
567.0 610.9 657.6 707.3 760.0 815.9 875.1 937.9 1004.4 1074.6
C
A P P E N D I X
THERMODYNAMIC QUANTITIES FOR SELECTED SUBSTANCES AT
298.15 K (25 °C) ¢Hf° (kJ/mol)
¢Gf° (kJ/mol)
S° (J/mol-K )
Aluminum Al(s) AlCl3(s) Al2O3(s)
0 -705.6 -1669.8
0 -630.0 -1576.5
28.32 109.3 51.00
Barium Ba(s) BaCO3(s) BaO(s)
0 -1216.3 -553.5
0 -1137.6 -525.1
63.2 112.1 70.42
Beryllium Be(s) BeO(s) Be(OH)2(s)
0 -608.4 -905.8
0 -579.1 -817.9
9.44 13.77 50.21
Bromine Br(g) Br -(aq) Br2(g) Br2(l) HBr(g)
111.8 -120.9 30.71 0 -36.23
82.38 -102.8 3.14 0 -53.22
174.9 80.71 245.3 152.3 198.49
145.5 0 -1128.76 -748.1 -1167.3 -604.17 -898.5 -1321.8
154.8 41.4 92.88 104.6 68.87 39.75 83.4 106.7
672.9 2.84 0 -64.0 -68.6 -635.1 -50.8 209.2
158.0 2.43 5.69 309.4 214.4 262.3 186.3 200.8
Substance
Calcium Ca(g) Ca(s) CaCO3(s, calcite) CaCl2(s) CaF2(s) CaO(s) Ca(OH)2(s) CaSO4(s) Carbon C(g) C(s, diamond) C(s, graphite) CCl4(g) CCl4(l) CF4(g) CH4(g) C2H2(g)
179.3 0 -1207.1 -795.8 -1219.6 -635.5 -986.2 -1434.0 718.4 1.88 0 -106.7 -139.3 -679.9 -74.8 226.77
Substance C2H4(g) C2H6(g) C3H8(g) C4H10(g) C4H10(l) C6H6(g) C6H6(l) CH3OH(g) CH3OH(l) C2H5OH(g) C2H5OH(l) C6H12O6(s) CO(g) CO2(g) CH3COOH(l)
¢Hf° (kJ/mol)
¢Gf° (kJ/mol)
S° (J/mol-K )
52.30 -84.68 -103.85 -124.73 -147.6 82.9 49.0 -201.2 -238.6 -235.1 -277.7 -1273.02 -110.5 -393.5 -487.0
68.11 -32.89 -23.47 -15.71 -15.0 129.7 124.5 -161.9 -166.23 -168.5 -174.76 -910.4 -137.2 -394.4 -392.4
219.4 229.5 269.9 310.0 231.0 269.2 172.8 237.6 126.8 282.7 160.7 212.1 197.9 213.6 159.8
Cesium Cs(g) Cs(l) Cs(s) CsCl(s)
76.50 2.09 0 -442.8
49.53 0.03 0 -414.4
175.6 92.07 85.15 101.2
Chlorine Cl(g) Cl1aq2 Cl2(g) HCl(aq) HCl(g)
121.7 -167.2 0 -167.2 -92.30
105.7 -131.2 0 -131.2 -95.27
165.2 56.5 222.96 56.5 186.69
Chromium Cr(g) Cr(s) Cr2O3(s)
397.5 0 -1139.7
352.6 0 -1058.1
174.2 23.6 81.2
Cobalt Co(g) Co(s)
439 0
393 0
179 28.4
Copper Cu(g) Cu(s)
338.4 0
298.6 0
166.3 33.30
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1060
APPENDIX C Thermodynamic Quantities for Selected Substances at 298.15 K (25 °C)
Substance
¢Hf° (kJ/mol)
¢Gf° (kJ/mol)
¢Hf° (kJ/mol)
¢Gf° (kJ/mol)
CuCl2(s) CuO(s) Cu2O(s)
-205.9 -156.1 -170.7
-161.7 -128.3 -147.9
108.1 42.59 92.36
MgO(s) Mg(OH)2(s)
-601.8 -924.7
-569.6 -833.7
26.8 63.24
158.7 - 13.8 202.7 173.51
Manganese Mn(g) Mn(s) MnO(s) MnO2(s) MnO4-1aq2
80.0 -332.6 0 -268.61
61.9 -278.8 0 -270.70
280.7 0 -385.2 -519.6 -541.4
238.5 0 -362.9 -464.8 -447.2
173.6 32.0 59.7 53.14 191.2
217.94 0 1536.2 0
203.26 0 1517.0 0
114.60 0 108.9 130.58
Mercury Hg(g) Hg(l) HgCl2(s) Hg2Cl2(s)
60.83 0 -230.1 -264.9
31.76 0 -184.0 -210.5
174.89 77.40 144.5 192.5
106.60 -55.19 62.25 0 25.94
70.16 -51.57 19.37 0 1.30
180.66 111.3 260.57 116.73 206.3
Nickel Ni(g) Ni(s) NiCl2(s) NiO(s)
429.7 0 -305.3 -239.7
384.5 0 -259.0 -211.7
182.1 29.9 97.65 37.99
415.5 0 -87.86 -47.69 -341.8 -400 -271.9 -822.16 -1117.1 -171.5
369.8 0 -84.93 -10.54 -302.3 -334 -255.2 -740.98 -1014.2 -160.1
180.5 27.15 113.4 293.3 117.9 142.3 60.75 89.96 146.4 52.92
Lead Pb(s) PbBr2(s) PbCO3(s) Pb(NO3)2(aq) Pb(NO3)2(s) PbO(s)
0 -277.4 -699.1 -421.3 -451.9 -217.3
0 -260.7 -625.5 -246.9 — -187.9
68.85 161 131.0 303.3 — 68.70
Nitrogen N(g) N2(g) NH3(aq) NH3(g) NH4 + 1aq2 N2H4(g) NH4CN(s) NH4Cl(s) NH4NO3(s) NO(g) NO2(g) N2O(g) N2O4(g) NOCl(g) HNO3(aq) HNO3(g)
472.7 0 -80.29 -46.19 -132.5 95.40 0.0 -314.4 -365.6 90.37 33.84 81.6 9.66 52.6 -206.6 -134.3
455.5 0 -26.50 -16.66 -79.31 159.4 — -203.0 -184.0 86.71 51.84 103.59 98.28 66.3 -110.5 -73.94
153.3 191.50 111.3 192.5 113.4 238.5 — 94.6 151 210.62 240.45 220.0 304.3 264 146 266.4
Lithium Li(g) Li(s) Li + 1aq2 Li + 1g2 LiCl(s)
159.3 0 -278.5 685.7 -408.3
126.6 0 -273.4 648.5 -384.0
138.8 29.09 12.2 133.0 59.30
Magnesium Mg(g) Mg(s) MgCl2(s)
Oxygen O(g) O2(g) O3(g) OH-1aq2 H2O(g) H2O(l) H2O2(g) H2O2(l)
247.5 0 142.3 -230.0 -241.82 -285.83 -136.10 -187.8
230.1 0 163.4 -157.3 -228.57 -237.13 -105.48 -120.4
161.0 205.0 237.6 - 10.7 188.83 69.91 232.9 109.6
147.1 0 -641.6
112.5 0 -592.1
148.6 32.51 89.6
Phosphorus P(g) P2(g)
316.4 144.3
280.0 103.7
163.2 218.1
Fluorine F(g) F1aq2 F2(g) HF(g) Hydrogen H(g) H + 1aq2 H + 1g2 H2(g) Iodine I(g) I -1aq2 I2(g) I2(s) HI(g) Iron Fe(g) Fe(s) Fe 2 + 1aq2 Fe 3 + 1aq2 FeCl2(s) FeCl3(s) FeO(s) Fe2O3(s) Fe3O4(s) FeS2(s)
S° (J/mol-K )
Substance
S° (J/mol-K )
APPENDIX C Thermodynamic Quantities for Selected Substances at 298.15 K 25 °C
Substance P4(g) P4(s, red) P4(s, white) PCl3(g) PCl3(l) PF5(g) PH3(g) P4O6(s) P4O10(s) POCl3(g) POCl3(l) H3PO4(aq)
¢Hf° (kJ/mol)
¢Gf° (kJ/mol)
S° (J/mol-K )
58.9 -17.46 0 -288.07 -319.6 -1594.4 5.4 -1640.1 -2940.1 -542.2 -597.0 -1288.3
24.4 -12.03 0 -269.6 -272.4 -1520.7 13.4 — -2675.2 -502.5 -520.9 -1142.6
280 22.85 41.08 311.7 217 300.8 210.2 — 228.9 325 222 158.2
Potassium K(g) K(s) KCl(s) KClO3(s) KClO3(aq) K2CO3(s) KNO3(s) K2O(s) KO2(s) K2O2(s) KOH(s) KOH(aq)
89.99 0 -435.9 -391.2 -349.5 -1150.18 -492.70 -363.2 -284.5 -495.8 -424.7 -482.4
61.17 0 -408.3 -289.9 -284.9 -1064.58 -393.13 -322.1 -240.6 -429.8 -378.9 -440.5
160.2 64.67 82.7 143.0 265.7 155.44 132.9 94.14 122.5 113.0 78.91 91.6
Rubidium Rb(g) Rb(s) RbCl(s) RbClO3(s)
85.8 0 -430.5 -392.4
55.8 0 -412.0 -292.0
170.0 76.78 92 152
377.8 0
336.1 0
Scandium Sc(g) Sc(s) Selenium H2Se(g) Silicon Si(g) Si(s) SiC(s) SiCl4(l) SiO2(s, quartz) Silver Ag(s) Ag + 1aq2 AgCl(s) Ag2O(s)
29.7 368.2 0 -73.22 -640.1 -910.9 0 105.90 -127.0 -31.05
15.9 323.9 0 -70.85 -572.8 -856.5 0 77.11 -109.70 -11.20
174.7 34.6 219.0 167.8 18.7 16.61 239.3 41.84 42.55 73.93 96.11 121.3
¢Gf° (kJ/mol)
1061
Substance
¢Hf° (kJ/mol)
AgNO3(s)
-124.4
Sodium Na(g) Na(s) Na + 1aq2 Na + 1g2 NaBr(aq) NaBr(s) Na2CO3(s) NaCl(aq) NaCl(g) NaCl(s) NaHCO3(s) NaNO3(aq) NaNO3(s) NaOH(aq) NaOH(s) Na2SO4(s)
107.7 0 -240.1 609.3 -360.6 -361.4 -1130.9 -407.1 -181.4 -410.9 -947.7 -446.2 -467.9 -469.6 -425.6 -1387.1
77.3 0 -261.9 574.3 -364.7 -349.3 -1047.7 -393.0 -201.3 -384.0 -851.8 -372.4 -367.0 -419.2 -379.5 -1270.2
153.7 51.45 59.0 148.0 141.00 86.82 136.0 115.5 229.8 72.33 102.1 207 116.5 49.8 64.46 149.6
Strontium SrO(s) Sr(g)
-592.0 164.4
-561.9 110.0
54.9 164.6
Sulfur S(s, rhombic) S8(g) SO2(g) SO3(g) SO42-1aq2 SOCl2(l) H2S(g) H2SO4(aq) H2SO4(l)
0 102.3 -296.9 -395.2 -909.3 -245.6 -20.17 -909.3 -814.0
0 49.7 -300.4 -370.4 -744.5 — -33.01 -744.5 -689.9
31.88 430.9 248.5 256.2 20.1 — 205.6 20.1 156.1
Titanium Ti(g) Ti(s) TiCl4(g) TiCl4(l) TiO2(s)
468 0 -763.2 -804.2 -944.7
422 0 -726.8 -728.1 -889.4
180.3 30.76 354.9 221.9 50.29
Vanadium V(g) V(s)
514.2 0
453.1 0
Zinc Zn(g) Zn(s) ZnCl2(s) ZnO(s)
130.7 0 -415.1 -348.0
95.2 0 -369.4 -318.2
-33.41
S° (J/mol-K ) 140.9
182.2 28.9
160.9 41.63 111.5 43.9
A P P E N D I X
D
AQUEOUS EQUILIBRIUM CONSTANTS
TABLE D.1 • Dissociation Constants for Acids at 25 ˚C Name
Formula
Ka1
Acetic acid Arsenic acid Arsenous acid Ascorbic acid Benzoic acid Boric acid Butanoic acid Carbonic acid Chloroacetic acid Chlorous acid Citric acid Cyanic acid Formic acid Hydroazoic acid Hydrocyanic acid Hydrofluoric acid Hydrogen chromate ion Hydrogen peroxide Hydrogen selenate ion Hydrogen sulfide Hypobromous acid Hypochlorous acid Hypoiodous acid Iodic acid Lactic acid Malonic acid Nitrous acid Oxalic acid Paraperiodic acid Phenol Phosphoric acid Propionic acid Pyrophosphoric acid Selenous acid Sulfuric acid Sulfurous acid Tartaric acid
CH3COOH (or HC2H3O2) H3AsO4 H3AsO3 H2C6H6O6 C6H5COOH (or HC7H5O2) H3BO3 C3H7COOH (or HC4H7O2) H2CO3 CH2ClCOOH (or HC2H2O2Cl) HClO2 HOOCC(OH) (CH2COOH)2 (or H3C6H5O7) HCNO HCOOH (or HCHO2) HN3 HCN HF HCrO4 -
1.8 * 10-5 5.6 * 10-3 5.1 * 10-10 8.0 * 10-5 6.3 * 10-5 5.8 * 10-10 1.5 * 10-5 4.3 * 10-7 1.4 * 10-3 1.1 * 10-2 7.4 * 10-4 3.5 * 10-4 1.8 * 10-4 1.9 * 10-5 4.9 * 10-10 6.8 * 10-4 3.0 * 10-7 2.4 * 10-12 2.2 * 10-2 9.5 * 10-8 2.5 * 10-9 3.0 * 10-8 2.3 * 10-11 1.7 * 10-1 1.4 * 10-4 1.5 * 10-3 4.5 * 10-4 5.9 * 10-2 2.8 * 10-2 1.3 * 10-10 7.5 * 10-3 1.3 * 10-5 3.0 * 10-2 2.3 * 10-3 Strong acid 1.7 * 10-2 1.0 * 10-3
1062
H2O2 HSeO4 H2S HBrO HClO HIO HIO3 CH3CH(OH)COOH (or HC3H5O3) CH2(COOH)2 (or H2C3H2O4) HNO2 (COOH)2 (or H2C2O4) H5IO6 C6H5OH (or HC6H5O) H3PO4 C2H5COOH (or HC3H5O2) H4P2O7 H2SeO3 H2SO4 H2SO3 HOOC(CHOH)2COOH (or H2C4H4O6)
Ka2
Ka3
1.0 * 10-7
3.0 * 10-12
1.6 * 10-12
5.6 * 10-11
1.7 * 10-5
4.0 * 10-7
1 * 10-19
2.0 * 10-6 6.4 * 10-5 5.3 * 10-9 6.2 * 10-8 4.4 5.3 1.2 6.4
* * * *
10-3 10-9 10-2 10-8
4.2 * 10-13 2.1 * 10-7
APPENDIX D Aqueous Equilibrium Constants
TABLE D.2 • Dissociation Constants for Bases at 25 ˚C Name
Formula
Kb
Ammonia Aniline Dimethylamine Ethylamine Hydrazine Hydroxylamine Methylamine Pyridine Trimethylamine
NH3 C6H5NH2 (CH3)2NH C2H5NH2 H2NNH2 HONH2 CH3NH2 C5H5N (CH3)3N
1.8 4.3 5.4 6.4 1.3 1.1 4.4 1.7 6.4
* * * * * * * * *
10-5 10-10 10-4 10-4 10-6 10-8 10-4 10-9 10-5
TABLE D.3 • Solubility-Product Constants for Compounds at 25 ˚C Name
Formula
Ksp
Name
Formula
Ksp
Barium carbonate Barium chromate Barium fluoride Barium oxalate Barium sulfate Cadmium carbonate Cadmium hydroxide Cadmium sulfide* Calcium carbonate (calcite) Calcium chromate Calcium fluoride Calcium hydroxide Calcium phosphate Calcium sulfate Chromium(III) hydroxide Cobalt(II) carbonate Cobalt(II) hydroxide Cobalt(II) sulfide* Copper(I) bromide Copper(II) carbonate Copper(II) hydroxide Copper(II) sulfide* Iron(II) carbonate Iron(II) hydroxide Lanthanum fluoride Lanthanum iodate Lead(II) carbonate Lead(II) chloride Lead(II) chromate
BaCO3 BaCrO4 BaF2 BaC2O4 BaSO4 CdCO3 Cd(OH)2 CdS CaCO3 CaCrO4 CaF2 Ca(OH)2 Ca3(PO4)2 CaSO4 Cr(OH)3 CoCO3 Co(OH)2 CoS CuBr CuCO3 Cu(OH)2 CuS FeCO3 Fe(OH)2 LaF3 La(IO3)3 PbCO3 PbCl2 PbCrO4
5.0 * 10-9 2.1 * 10-10 1.7 * 10-6 1.6 * 10-6 1.1 * 10-10 1.8 * 10-14 2.5 * 10-14 8 * 10-28 4.5 * 10-9 4.5 * 10-9 3.9 * 10-11 6.5 * 10-6 2.0 * 10-29 2.4 * 10-5 1.6 * 10-30 1.0 * 10-10 1.3 * 10-15 5 * 10-22 5.3 * 10-9 2.3 * 10-10 4.8 * 10-20 6 * 10-37 2.1 * 10-11 7.9 * 10-16 2 * 10-19 7.4 * 10-14 7.4 * 10-14 1.7 * 10-5 2.8 * 10-13
Lead(II) fluoride Lead(II) sulfate Lead(II) sulfide* Magnesium hydroxide Magnesium carbonate Magnesium oxalate Manganese(II) carbonate Manganese(II) hydroxide Manganese(II) sulfide* Mercury(I) chloride Mercury(I) iodide Mercury(II) sulfide* Nickel(II) carbonate Nickel(II) hydroxide Nickel(II) sulfide* Silver bromate Silver bromide Silver carbonate Silver chloride Silver chromate Silver iodide Silver sulfate Silver sulfide* Strontium carbonate Tin(II) sulfide* Zinc carbonate Zinc hydroxide Zinc oxalate Zinc sulfide*
PbF2 PbSO4 PbS Mg(OH)2 MgCO3 MgC2O4 MnCO3 Mn(OH)2 MnS Hg2Cl2 Hg2I2 HgS NiCO3 Ni(OH)2 NiS AgBrO3 AgBr Ag2CO3 AgCl Ag2CrO4 AgI Ag2SO4 Ag2S SrCO3 SnS ZnCO3 Zn(OH)2 ZnC2O4 ZnS
3.6 * 10-8 6.3 * 10-7 3 * 10-28 1.8 * 10-11 3.5 * 10-8 8.6 * 10-5 5.0 * 10-10 1.6 * 10-13 2 * 10-53 1.2 * 10-18 1.1 * 10-1.1 2 * 10-53 1.3 * 10-7 6.0 * 10-16 3 * 10-20 5.5 * 10-13 5.0 * 10-13 8.1 * 10-12 1.8 * 10-10 1.2 * 10-12 8.3 * 10-17 1.5 * 10-5 6 * 10-51 9.3 * 10-10 1 * 10-26 1.0 * 10-10 3.0 * 10-16 2.7 * 10-8 2 * 10-25
*For a solubility equilibrium of the type MS(s) + H2O(l) Δ M2 + (aq) + HS-(aq) + OH-(aq)
1063
A P P E N D I X
E
STANDARD REDUCTION POTENTIALS AT
25 °C
Half-Reaction +
E°(V)
-
Ag (aq) + e ¡ Ag(s) AgBr(s) + e - ¡ Ag(s) + Br -(aq) AgCl(s) + e - ¡ Ag(s) + Cl-(aq) Ag(CN)2 -(aq) + e - ¡ Ag(s) + 2 CN -(aq) Ag 2CrO4(s) + 2 e - ¡ 2 Ag(s) + CrO42-(aq) AgI(s) + e - ¡ Ag(s) + I -(aq) Ag(S2O3)23-(aq) + e - ¡ Ag(s) + 2 S2O32-(aq) Al3 + (aq) + 3 e - ¡ Al(s) H3AsO4(aq) + 2 H + (aq) + 2 e - ¡ H3AsO3(aq) + H2O(l) Ba2 + (aq) + 2 e - ¡ Ba(s) BiO + (aq) + 2 H + (aq) + 3 e - ¡ Bi(s) + H2O(l) Br2(l) + 2 e - ¡ 2 Br -(aq) 2 BrO3 -(aq) + 12 H + (aq) + 10 e - ¡ Br2(l) + 6 H2O(l) 2 CO2(g) + 2 H + (aq) + 2 e - ¡ H2C2O4(aq) Ca2 + (aq) + 2 e - ¡ Ca(s) Cd2 + (aq) + 2 e - ¡ Cd(s) Ce 4 + (aq) + e - ¡ Ce 3 + (aq) Cl 2(g) + 2 e - ¡ 2 Cl-(aq) 2 HClO(aq) + 2 H + (aq) + 2 e - ¡ ClO-(aq) + H2O(l) + 2 e - ¡
Cl 2(g) + 2 H2O(l)
Cl-(aq) + 2 OH-(aq) 2 ClO3 (aq) + 12 H (aq) + 10 e ¡ Cl 2(g) + 6 H2O(l) Co2 + (aq) + 2 e - ¡ Co(s) Co3 + (aq) + e - ¡ Co2 + (aq) Cr 3 + (aq) + 3 e - ¡ Cr(s) Cr 3 + (aq) + e - ¡ Cr 2 + (aq) CrO72-(aq) + 14 H + (aq) + 6 e - ¡ 2 Cr 3 + (aq) + 7 H2O(l) 2CrO4 (aq) + 4 H2O(l) + 3 e ¡ Cr(OH)3(s) + 5 OH-(aq) 2+ Cu (aq) + 2 e ¡ Cu(s) Cu2 + (aq) + e - ¡ Cu + (aq) Cu + (aq) + e - ¡ Cu(s) CuI(s) + e - ¡ Cu(s) + I -(aq) F2(g) + 2 e - ¡ 2 F -(aq) -
+
-
Fe 2 + (aq) + 2 e - ¡ Fe(s) Fe 3 + (aq) + e - ¡ Fe 2 + (aq) Fe(CN)63-(aq) + e - ¡ Fe(CN)64-(aq) 2 H + (aq) + 2 e - ¡ H2(g)
1064
+0.799 +0.095 +0.222 -0.31 +0.446 -0.151 +0.01 -1.66 +0.559 -2.90 +0.32 +1.065 +1.52 -0.49 -2.87 -0.403 +1.61 +1.359 +1.63 +0.89 +1.47 -0.277 +1.842 -0.74 -0.41 +1.33 -0.13 +0.337 +0.153 +0.521 -0.185 +2.87 -0.440 +0.771 +0.36 0.000
Half-Reaction
E°(V)
-
-
2 H2O(l) + 2 e ¡ H2(g) + 2 OH (aq) HO2 -(aq) + H2O(l) + 2 e - ¡ 3 OH-(aq) H2O2(aq) + 2 H + (aq) + 2 e - ¡ 2 H2O(l) Hg 22 + (aq) + 2 e - ¡ 2 Hg(l) 2 Hg2 + (aq) + 2 e - ¡ Hg 22 + (aq) Hg2 + (aq) + 2 e - ¡ Hg(l) I2(s) + 2 e - ¡ 2 I -(aq) 2 IO3 -(aq) + 12 H + (aq) + 10 e - ¡ K + (aq) + e - ¡ K(s) Li + (aq) + e - ¡ Li(s) Mg 2 + (aq) + 2 e - ¡ Mg(s) Mn2 + (aq) + 2 e - ¡ Mn(s) MnO2(s) + 4 H + (aq) + 2 e - ¡
I2(s) + 6 H2O(l)
Mn2 + (aq) + 2 H2O(l) MnO4 (aq) + 8 H (aq) + 5 e ¡ Mn2 + (aq) + 4 H2O(l) MnO4 (aq) + 2 H2O(l) + 3 e ¡ MnO2(s) + 4 OH-(aq) -
+
-
HNO2(aq) + H + (aq) + e - ¡ NO(g) + H2O(l) N2(g) + 4 H2O(l) + 4 e - ¡ 4 OH-(aq) + N2H4(aq) N2(g) + 5 H + (aq) + 4 e - ¡ N2H5 + (aq) NO3 -(aq) + 4 H + (aq) + 3 e - ¡ NO(g) + 2 H2O(l) Na + (aq) + e - ¡ Na(s) Ni2 + (aq) + 2 e - ¡ Ni(s) O2(g) + 4 H + (aq) + 4 e - ¡ 2 H2O(l) O2(g) + 2 H2O(l) + 4 e - ¡ 4 OH-(aq) O2(g) + 2 H + (aq) + 2 e - ¡ H2O2(aq) O3(g) + 2 H + (aq) + 2 e - ¡ O2(g) + H2O(l) Pb 2 + (aq) + 2 e - ¡ Pb(s) PbO2(s) + HSO4 -(aq) + 3 H + (aq) + 2 e - ¡ PbSO4(s) + 2 H2O(l) PbSO4(s) + H + (aq) + 2 e - ¡ Pb(s) + HSO4 -(aq) PtCl 42-(aq) + 2 e - ¡ Pt(s) + 4 Cl-(aq) S(s) + 2 H + (aq) + 2 e - ¡ H2S(g) H2SO3(aq) + 4 H + (aq) + 4 e - ¡ S(s) + 3 H2O(l) HSO4 -(aq) + 3 H + (aq) + 2 e - ¡ H2SO3(aq) + H2O(l) Sn2 + (aq) + 2 e - ¡ Sn(s) Sn4 + (aq) + 2 e - ¡ Sn2 + (aq) VO2 + (aq) + 2 H + (aq) + e - ¡ VO2 + (aq) + H2O(l) Zn2 + (aq) + 2 e - ¡ Zn(s)
-0.83 +0.88 +1.776 +0.789 +0.920 +0.854 +0.536 +1.195 -2.925 -3.05 -2.37 -1.18 +1.23 +1.51 +0.59 +1.00 -1.16 -0.23 +0.96 -2.71 -0.28 +1.23 +0.40 +0.68 +2.07 -0.126 +1.685 -0.356 +0.73 +0.141 +0.45 +0.17 -0.136 +0.154 +1.00 -0.763
ANSWERS TO SELECTED EXERCISES CHAPTER 1 1.1 (a) Pure element: i (b) mixture of elements: v, vi (c) pure compound: iv (d) mixture of an element and a compound: ii, iii 1.3 This kind of separation based on solubility differences is called extraction. The insoluble grounds are then separated from the coffee solution by filtration. 1.5 (a) The aluminum sphere is lightest, then nickel, then silver. (b) The platinum sphere is largest, then gold, then lead. 1.7 (a) 7.5 cm; two significant figures (sig figs) (b) 72 mi/hr (inner scale, two significant figures) or 115 km/hr (outer scale, three significant figures) 1.9 Arrange the conversion factor so that the given unit cancels and the desired unit is in the correct position. 1.11 (a) Heterogeneous mixture (b) homogeneous mixture (heterogeneous if there are undissolved particles) (c) pure substance (d) pure substance. 1.13 (a) S (b) Au (c) K (d) Cl (e) Cu (f) uranium (g) nickel (h) sodium (i) aluminum (j) silicon 1.15 C is a compound; it contains both carbon and oxygen. A is a compound; it contains at least carbon and oxygen. B is not defined by the data given; it is probably also a compound because few elements exist as white solids. 1.17 Physical properties: silvery white; lustrous; melting point = 649 °C; boiling point = 1105 °C; density at 20 °C = 1.738 g>cm3; pounded into sheets; drawn into wires; good conductor. Chemical properties: burns in air; reacts with Cl2. 1.19 (a) Chemical (b) physical (c) physical (d) chemical (e) chemical 1.21 (a) Add water to dissolve the sugar; filter this mixture, collecting the sand on the filter paper and the sugar water in the flask. Evaporate water from the flask to recover solid sugar. (b) Allow the mixture to settle so that there are two distinct layers. Carefully pour off most of the top oil layer. After the layers reform, use a dropper to remove any remaining oil. Vinegar is in the original vessel and oil is in a second container. 1.23 (a) 1 * 10-1 (b) 1 * 10-2 (c) 1 * 10-15 (d) 1 * 10-6 (e) 1 * 106 (f) 1 * 103 (g) 1 * 10-9 (h) 1 * 10-3 (i) 1 * 10-12 1.25 (a) 22 °C (b) 422.1 °F (c) 506 K (d) 107 °C (e) 1644 K (f) -459.67 °F 1.27 (a) 1.62 g/mL. Tetrachloroethylene, 1.62 g/mL, is more dense than water, 1.00 g/mL; tetrachloroethylene will sink rather than float on water. (b) 11.7 g 1.29 (a) Calculated density = 0.86 g>mL. The substance is probably toluene, density = 0.866 g>mL. (b) 40.4 mL ethylene glycol (c) 1.11 * 103 g nickel 1.31 28 Pg 1.33 Exact: (c), (d), and (f) 1.35 (a) 3 (b) 2 (c) 5 (d) 3 (e) 5 (f) 1 1.37 (a) 1.025 * 102 (b) 6.570 * 105 (c) 8.543 * 10-3 (d) 2.579 * 10-4 (e) -3.572 * 10-2 1.39 (a) 17.00 (b) 812.0 (c) 8.23 * 103 (d) 8.69 * 10-2 1.41 5 significant figures 1 * 10-3 g 1 kg 1 * 10-3 m 1 nm 1.43 (a) (b) * * -9 1 mm 1 mg 1000 g 1 * 10 m (2.54)3 cm3 1 in. 1 cm 1 ft 1000 m (d) * * * 1 km 2.54 cm 12 in. 1 * 10-2 m 13 in.3 3 1.45 (a) 54.7 km/hr (b) 1.3 * 10 gal (c) 46.0 m (d) 0.984 in/hr 1.47 (a) 4.32 * 105 s (b) 88.5 m (c) $0.499/L (d) 46.6 km/hr (e) 1.420 L/s (f) 707.9 cm3 1.49 (a) 1.2 * 102 L (b) 5 * 102 mg (c) 19.9 mi/gal (2 * 101 mi/gal for 1 significant figure) (d) 26 mL/g (3 * 101 mL/g for 1 significant figure) 1.51 64 kg air 1.53 14-in. shoe 6 57-cm string 6 1.1-m pipe 1.55 $6 * 104 1.59 8.47 g O; the law of constant composition 1.62 (a) Volume (b) area (c) volume (d) density (e) time (f) length (g) temperature 1.65 (a) 1.13 * 105 quarters (b) 6.41 * 105 g (c) $2.83 * 104 (d) 4.13 * 108 stacks 1.68 The most dense liquid, Hg, will sink; the least dense, cyclohexane, will float; H2O will be in the middle. 1.71 density of solid = 1.63 g>mL 1.74 (a) Density of peat = 0.13 g>cm3, density of soil = 2.5 g>cm3. It is not correct to say that peat is “lighter” than topsoil. Volumes must be specified in order to compare masses. (b) Buy 16 bags of peat (more than 15 are needed). (Results to 1 significant figure are not meaningful.) 1.77 The inner diameter of the tube is 1.13 cm. 1.79 The separation (c)
is successful if two distinct spots are seen on the paper. To quantify the characteristics of the separation, calculate a reference value for each spot: distance traveled by spot/distance traveled by solvent. If the values for the two spots are fairly different, the separation is successful.
CHAPTER 2 2.1 (a) The path of the charged particle bends because the particle is repelled by the negatively charged plate and attracted to the positively charged plate. (b) (-) (c) increase (d) decrease 2.4 The particle is 2an ion. 32 2.6 Formula: IF5; name: iodine pentafluoride; the 16S compound is molecular. 2.9 Postulate 4 of the atomic theory states that the relative number and kinds of atoms in a compound are constant, regardless of the source. Therefore, 1.0 g of pure water should always contain the same relative amounts of hydrogen and oxygen, no matter where or how the sample is obtained. 2.11 (a) 0.5711 g O>1 g N; 1.142 g O>1 g N; 2.284 g O>1 g N; 2.855 g O>1 g N (b) The numbers in part (a) obey the law of multiple proportions. Multiple proportions arise because atoms are the indivisible entities combining, as stated in Dalton’s atomic theory. 2.13 (1) Electric and magnetic fields deflected the rays in the same way they would deflect negatively charged particles. (2) A metal plate exposed to cathode rays acquired a negative charge. 2.15 (a) Most of the volume of an atom is empty space in which electrons move. Most alpha particles passed through this space. (b) The few alpha particles that hit the massive, positively charged gold nuclei were strongly repelled and deflected back in the direction they came from. (c) Because the Be nuclei have a smaller volume and a smaller positive charge than the Au nuclei, fewer alpha particles will be scattered and fewer will be strongly back scattered. 2.17 (a) 0.135 nm; 1.35 * 102 or 135 pm (b) 3.70 * 106 Au atoms (c) 1.03 * 10-23 cm3 2.19 (a) Proton, neutron, electron (b) proton = 1+ , neutron = 0, electron = 1- (c) The neutron is most massive. (The neutron and proton have very similar masses.) (d) The electron is least massive. 2.21 (a) Atomic number is the number of protons in the nucleus of an atom. Mass number is the total number of nuclear particles, protons plus neutrons, in an atom. (b) mass number 2.23 (a) 40Ar: 18 p, 22 n, 18 e (b) 65Zn: 30 p, 35 n, 30 e (c) 70Ga: 31 p, 39 n, 31 e (d) 80Br: 35 p, 45 n, 35 e (e) 184W: 74 p, 110 n, 74 e (f) 243Am: 95 p, 148 n, 95e 2.25 Symbol
52
55
112
222
Protons Neutrons Electrons Mass no.
24 28 24 52
25 30 25 55
48 64 48 112
86 136 86 222
Cr
Mn
Cd
Rn
207
Pb
82 125 82 207
84 75 24 12 2.27 (a) 196 78Pt (b) 36Kr (c) 33As (d) 12Mg 2.29 (a) 6 C (b) Atomic weights are average atomic masses, the sum of the mass of each naturally occurring isotope of an element times its fractional abundance. Each B atom will have the mass of one of the naturally occurring isotopes, while the “atomic weight” is an average value. 2.31 63.55 amu 2.33 (a) In Thomson’s cathode-ray experiments and in mass spectrometry, a stream of charged particles is passed through the poles of a magnet. The charged particles are deflected by the magnetic field according to their mass and charge. (b) The x-axis label is atomic weight, and the y-axis label is signal intensity. (c) Uncharged particles are not deflected in a magnetic field. The effect of the magnetic field on charged moving particles is the basis of their separation by mass. 2.35 (a) average atomic mass = 24.31 amu
A-1
A-2
Answers to Selected Exercises
Signal intensity
(b) (7.9)
5
(1.1)
(1)
24
25
26
Atomic weight (amu) 2.37 (a) Cr, 24 (metal) (b) He, 2 (nonmetal) (c) P, 15 (nonmetal) (d) Zn, 30 (metal) (e) Mg, 12 (metal) (f) Br, 35 (nonmetal) (g) As, 33 (metalloid) 2.39 (a) K, alkali metals (metal) (b) I, halogens (nonmetal) (c) Mg, alkaline earth metals (metal) (d) Ar, noble gases (nonmetal) (e) S, chalcogens (nonmetal) 2.41 An empirical formula shows the simplest mole ratio of elements in a compound. A molecular formula shows the exact number and kinds of atoms in a molecule. A structural formula shows which atoms are attached to which. 2.43 (a) AlBr3 (b) C4H5 (c) C2H4O (d) P2O5 (e) C3H2Cl (f) BNH2 2.45 (a) 6 (b) 6 (c) 12
2.59 Molecular: (a) B2H6 (b) CH3OH (f) NOCl (g) NF3. Ionic: (c) LiNO3 (d) Sc2O3 (e) CsBr (h) Ag2SO4 2.61 (a) ClO2- (b) Cl(c) ClO3- (d) ClO4- (e) ClO- 2.63 (a) calcium, 2+ ; oxide, 2- (b) sodium, 1 + ; sulfate, 2- (c) potassium, 1+ ; perchlorate, 1- (d) iron, 2+ , nitrate, 1 - (e) chromium, 3 + ; hydroxide, 1 - 2.65 (a) lithium oxide (b) iron(III) chloride (ferric chloride) (c) sodium hypochlorite (d) calcium sulfite (e) copper(II) hydroxide (cupric hydroxide) (f) iron(II) nitrate (ferrous nitrate) (g) calcium acetate (h) chromium(III) carbonate (chromic carbonate) (i) potassium chromate (j) ammonium sulfate 2.67 (a) Al(OH)3 (b) K2SO4 (c) Cu2O (d) Zn(NO3)2 (e) HgBr2 (f) Fe2(CO3)3 (g) NaBrO 2.69 (a) Bromic acid (b) hydrobromic acid (c) phosphoric acid (d) HClO (e) HIO3 (f) H2SO3 2.71 (a) Sulfur hexafluoride (b) iodine pentafluoride (c) xenon trioxide (d) N2O4 (e) HCN (f) P4S6 2.73 (a) ZnCO3, ZnO, CO2 (b) HF, SiO2, SiF4, H2O (c) SO2, H2O, H2SO3 (d) PH3 (e) HClO4, Cd, Cd(ClO4)2 (f) VBr3 2.75 (a) A hydrocarbon is a compound composed of the elements hydrogen and carbon only. (b) H
(a) C2H6O,
H
C
H
O
C
H
(c) (b) C2H6O,
H
H
H
C
C
H
H
H O
H
C
2.79 (a, b)
O
H
H
H (d) PF3 ,
F
P
H
C
C
C
C
H
H
H
H
H
H
H
H
H
C
C
C
C
H
H
H
H
OH
H
H (c) CH4O,
H
2.77 (a) Functional groups are groups of specific atoms that are constant from one molecule to the next. (b) ¬ OH
H
H
H
Molecular: C4H10 Empirical: C2H5
2.47 H
H
H
H
H
C
C
C
H
H
H
Cl
1-chloropropane
F
F 2.49 Co3 +
Symbol
59
80
Se2-
Protons Neutrons Electrons Net Charge
27 32 24 3+
34 46 36 2-
Os2 +
Hg2 +
192
200
76 116 74 2+
80 120 78 2+
2.51 (a) Mg2 + (b) Al3 + (c) K + (d) S2- (e) F- 2.53 (a) GaF3, gallium(III) fluoride (b) LiH, lithium hydride (c) AlI3, aluminum iodide (d) K2S, potassium sulfide 2.55 (a) CaBr2 (b) K2CO3 (c) Al(CH3COO)3 (d) (NH4)2SO4 (e) Mg3(PO4)2 2.57 Ion
K+
NH4+
Mg2 +
Fe3 +
ClOH-
KCl KOH K2CO3 K3PO4
NH4Cl NH4OH (NH4)2CO3 (NH4)3PO4
MgCl2 Mg(OH)2 MgCO3 Mg3(PO4)2
FeCl3 Fe(OH)3 Fe2(CO3)3 FePO4
CO32PO43-
H
H
Cl
H
C
C
C
H
H
H
H
2-chloropropane 2.82 (a) 2 protons, 1 neutron, 2 electrons (b) tritium, 3H, is more massive. (c) A precision of 1 * 10-27 g would be required to differentiate between 3H + and 3He + . 2.84 Arrangement A, 4.1 * 1014 atoms>cm2 (b) Arrangement B, 4.7 * 1014 atoms>cm2 (c) The ratio of atoms going from arrangement B to arrangement A is 1.2 to 1. In three dimensions, arrangement B leads to a greater density for Rb metal. 2.87 (a) 168O, 178O, 18 8O (b) All isotopes are atoms of the same element, oxygen, with the same atomic number, 8 protons in the nucleus and 8 electrons. We expect their electron arrangements to be the same and their chemical properties to be very similar. Each has a different number of neutrons, a different mass number, and a different atomic mass. 2.90 (a) 69 31Ga, 69 31 protons, 38 neutrons; 71 31Ga, 31 protons, 40 neutrons (b) 31Ga, 60.3%, 71 31Ga, 39.7%. 2.93 (a) 5 significant figures (b) An electron is 0.05444% of the mass of an 1H atom. 2.96 Strontium is an alkaline earth metal, similar in chemical properties to calcium and magnesium. Harmful strontium closely mimics essential calcium and magnesium, then behaves badly when the body tries to use it as it uses calcium and
Answers to Selected Exercises magnesium. 2.98 (a) Nickel(II) oxide, 2 + (b) manganese(IV) oxide, 4+ (c) chromium(III) oxide, 3 + (d) molybdenum(VI) oxide, 6+ 2.101 (a) Perbromate ion (b) selenite ion (c) AsO43 - (d) HTeO42.104 (a) Potassium nitrate (b) sodium carbonate (c) calcium oxide (d) hydrochloric acid (e) magnesium sulfate (f) magnesium hydroxide
CHAPTER 3 3.1 Equation (a) best fits the diagram. 3.3 (a) NO2 (b) No, because we have no way of knowing whether the empirical and molecular formulas are the same. NO2 represents the simplest ratio of atoms in a molecule but not the only possible molecular formula. 3.5 (a) C2H5NO2 (b) 75.0 g>mol (c) 225 g glycine (d) Mass %N in glycine is 18.7%. 3.7
N2 + 3 H2 ¡ 2 NH3. Eight N atoms (4 N2 molecules) require 24 H atoms (12 H2 molecules) for complete reaction. Only 9 H2 molecules are available, so H2 is the limiting reactant. Nine H2 molecules (18 H atoms) determine that 6 NH3 molecules are produced. One N2 molecule is in excess. 3.9 (a) Conservation of mass (b) Subscripts in chemical formulas should not be changed when balancing equations, because changing the subscript changes the identity of the compound (law of constant composition). (c) H2O(l), H2O(g), NaCl(aq), NaCl(s) 3.11 (a) 2 CO(g) + O2(g) ¡ 2 CO2(g) (b) N2O5(g) + H2O(l) ¡ 2 HNO3(aq) (c) CH4(g) + 4 Cl2(g) ¡ CCl4(l) + 4 HCl(g) (d) Al4C3(s) + 12 H2O(l) ¡ 4 Al(OH)3(s) + 3 CH4(g) (e) 2 C5H10O2(l) + 13 O2(g) ¡ 10 CO2(g) + 10 H2O(g) (f) 2 Fe(OH)3(s) + 3 H2SO4(aq) ¡ Fe2(SO4)3(aq) + 6 H2O(l) (g) Mg3N2(s) + 4 H2SO4(aq) ¡ 3 MgSO4(aq) + (NH4)2SO4(aq) 3.13 (a) CaC2(s) + 2 H2O(l) ¡ Ca(OH)2(aq) + C2H2(g) ¢ (b) 2 KClO3(s) ¡ 2 KCl(s) + 3 O2(g) (c) Zn(s) + H2SO4(aq) ¡ ZnSO4(aq) + H2(g) (d) PCl3(l) + 3 H2O(l) ¡ H3PO3(aq) + 3 HCl(aq) (e) 3 H2S(g) + 2 Fe(OH)3(s) ¡ Fe2S3(s) + 6 H2O(g) 3.15 (a) Determine the formula by balancing the positive and negative charges in the ionic product. All ionic compounds are solids. 2 Na(s) + Br2(l) ¡ 2 NaBr(s) (b) The second reactant is O2(g). The products are CO2(g) and H2O(l). 2 C6H6(l) + 15 O2(g) ¡ 12 CO2(g) + 6 H2O(l) 3.17 (a) Mg(s) + Cl 2(g) ¡ MgCl 2(s) ¢
(b) BaCO3(s) ¡ BaO(s) + CO2(g) (c) C8H8(l) + 10 O2(g) ¡ 8 CO2(g) + 4 H2O(l) (d) C2H6O(g) + 3 O2(g) ¡ 2 CO2(g) + 3 H2O(l) 3.19 (a) 2 C3H6(g) + 9 O2(g) ¡ 6 CO2(g) + 6 H2O(g) combustion (b) NH4NO3(s) ¡ N2O(g) + 2 H2O(g) decomposition (c) C5H6O(l) + 6 O2(g) ¡ 5 CO2(g) + 3 H2O(g) combustion (d) N2(g) + 3 H2(g) ¡ 2 NH3(g) combination (e) K2O(s) + H2O(l) ¡ 2 KOH(aq) combination 3.21 (a) 63.0 amu (b) 158.0 amu (c) 310.3 amu (d) 60.1 amu (e) 235.7 amu (f) 392.3 amu (g) 137.5 amu 3.23 (a) 16.8% (b) 16.1% (c) 21.1% (d) 28.8% (e) 27.2% (f) 26.5% 3.25 (a) 79.2% (b) 63.2%
A-3
(c) 64.6% 3.27 (a) 6.022 * 1023 (b) The formula weight of a substance in amu has the same numerical value as the molar mass expressed in grams. 3.29 23 g Na contains 1 mol of atoms; 0.5 mol H2O contains 1.5 mol atoms; 6.0 * 1023 N2 molecules contain 2 mol of atoms. 3.31 4.37 * 1025 kg (assuming 160 lb has 3 significant figures). One mole of people weighs 7.31 times as much as Earth. 3.33 (a) 35.9 g C12H22O11 (b) 0.75766 mol Zn(NO3)2 (c) 6.0 * 1017 CH3CH2OH molecules (d) 2.47 * 1023 N atoms 3.35 (a) 0.373 g (NH4)3PO4 (b) 5.737 * 10-3 mol Cl- (c) 0.248 g C8H10N4O2 (d) 387 g cholesterol>mol 3.37 (a) Molar mass = 162.3 g (b) 3.08 * 10-5 mol allicin (c) 1.86 * 1019 allicin molecules (d) 3.71 * 1019 S atoms 3.39 (a) 2.500 * 1021 H atoms (b) 2.083 * 1020 C6H12O6 molecules (c) 3.460 * 10-4 mol C6H12O6 (d) 0.06227 g C6H12O6 3.41 3.2 * 10-8 mol C2H3Cl>L; 1.9 * 1016 molecules>L 3.43 (a) C2H6O (b) Fe2O3 (c) CH2O 3.45 (a) CSCl2 (b) C3OF6 (c) Na3AlF6 3.47 31 g/mol 3.49 (a) C6H12 (b) NH2Cl 3.51 (a) Empirical formula, CH; molecular formula, C8H8 (b) empirical formula, C4H5N2O; molecular formula, C8H10N4O2 (c) empirical formula and molecular formula, NaC5H8O4N 3.53 (a) C7H8 (b) The empirical and molecular formulas are C10H20O. 3.55 The molecular formula from the model is C8H16O2, which corresponds to an empirical formula of C4H8O. The results from combustion analysis are consistent with this empirical formula. 3.57 x = 10; Na2CO3 # 10 H2O 3.59 If the equation is not balanced, the mole ratios derived from the coefficients will be incorrect and lead to erroneous calculated amounts of products. 3.61 (a) 2.40 mol HF (b) 5.25 g NaF (c) 0.610 g Na2SiO3 3.63 (a) Al(OH)3(s) + 3 HCl(aq) ¡ AlCl3(aq) + 3 H2O(l) (b) 0.701 g HCl (c) 0.855 g AlCl3; 0.347 g H2O (d) Mass of reactants = 0.500 g + 0.701 g = 1.201 g; mass of products = 0.855 g + 0.347 g = 1.202 g. Mass is conserved, within the precision of the data. 3.65 (a) Al2S3(s) + 6 H2O(l) ¡ 2 Al(OH)3(s) + 3 H2S(g) (b) 14.7 g Al(OH)3 3.67 (a) 2.25 mol N2 (b) 15.5 g NaN3 (c) 548 g NaN3 3.69 (a) 5.50 * 10-3 mol Al (b) 1.47 g AlBr3 3.71 (a) The limiting reactant determines the maximum number of product moles resulting from a chemical reaction; any other reactant is an excess reactant. (b) The limiting reactant regulates the amount of products because it is completely used up during the reaction; no more product can be made when one of the reactants is unavailable. (c) Combining ratios are molecule and mole ratios. Since different molecules have different masses, comparing initial masses of reactants will not provide a comparison of numbers of molecules or moles. 3.73 (a) 2255 bicycles (b) 50 frames left over, 305 wheels left over (c) the handlebars 3.75 NaOH is the limiting reactant; 0.925 mol Na2CO3 can be produced; 0.075 mol CO2 remains. 3.77 (a) NaHCO3 is the limiting reactant. (b) 0.524 g CO2 (c) 0.238 g citric acid remains 3.79 0.00 g AgNO3 (limiting reactant), 1.94 g Na2CO3, 4.06 g Ag2CO3, 2.50 g NaNO3 3.81 (a) The theoretical yield is 60.3 g C6H5Br. (b) 70.1% yield 3.83 28 g S8 actual yield 3.85 (a) C2H4O2(l) + 2 O2(g) ¡ 2 CO2(g) + 2 H2O(l) (b) Ca(OH)2(s) ¡ CaO(s) + H2O(g) (c) Ni(s) + Cl2(g) ¡ NiCl2(s) 3.89 (a) 8 * 10-20 g Si (b) 2 * 103 Si atoms (with 2 significant figures, 1700 Si atoms) (c) 1 * 103 Ge atoms (with 2 significant figures, 1500 Ge atoms) 3.93 C8H8O3 3.97 (a) 1.19 * 10-5 mol NaI (b) 8.1 * 10-3 g NaI 3.101 7.5 mol H2 and 4.5 mol N2 present initially 3.105 6.46 * 1024 O atoms 3.107 (a) 88 kg CO2 (b) 4 * 102 (400) kg CO2 3.109 (a) S(s) + O2(g) ¡ SO2(g); SO2(g) + CaO(s) ¡ CaSO3(s) (b) 7.9 * 107 g CaO (c) 1.7 * 108 g CaSO3
CHAPTER 4 4.1 Diagram (c) represents Li2SO4 4.3 (a) HCOOH is a weak electrolyte. (b) HNO3 is a strong electrolyte. (c) CH3CH2OH is a nonelectrolyte. 4.5 BaCl2 4.7 (b) NO3- and (c) NH4 + will always be spectator ions. 4.9 In a redox reaction, electrons are transferred from the oxidized substance to the reduced substance. In an acid–base reaction, protons are transferred from an acid to a base. 4.11 No. Electrolyte solutions conduct electricity because the dissolved ions carry charge through the solution from one electrode to the other. 4.13 Although H2O molecules are electrically neutral, there is an unequal distribution of electrons throughout the molecule. The partially
A-4
Answers to Selected Exercises
positive ends of H2O molecules are attracted to anions in the solid, while the partially negative ends are attracted to cations. Thus, both cations and anions in an ionic solid are surrounded and separated (dissolved) by H2O. We do not expect ionic compounds to be soluble in molecular liquids such as Br2(l) or Hg(l). There is a symmetrical charge distribution in Hg atoms and Br2 molecules, so there are no attractive forces to stabilize the separated ions of an ionic solid. 4.15 (a) ZnCl 2(aq) ¡ Zn2 + (aq) + 2 Cl-(aq) (b) HNO3(aq) ¡ (NH4)2SO4(aq) ¡ 2 NH4 + (aq) + H + (aq) + NO3-(aq) (c) 2SO4 (aq) (d) Ca(OH)2(aq) ¡ Ca2 + (aq) + 2 OH-(aq) 4.17 HCOOH molecules, H + ions , and HCOO- ions ; HCOOH(aq) Δ H + (aq) + HCOO - (aq) 4.19 (a) Soluble (b) insoluble (c) soluble (d) soluble (e) soluble 4.21 (a)Na2CO3(aq) + 2 AgNO3(aq) ¡ Ag2CO3(s) + 2 NaNO3(aq) (b) No precipitate (c) FeSO4(aq) + Pb(NO3)2(aq) ¡ PbSO4(s) + Fe(NO3)2(aq) 4.23 (a) Na + , SO42- (b) Na + , NO3- (c) NH4 + , Cl4.25 The solution contains Pb2 + . 4.27 Compound
Ba(NO3)2 Result
NaCl Result
AgNO3(aq) CaCl2(aq) Al2(SO4)3(aq)
No ppt No ppt BaSO4 ppt
AgCl ppt No ppt No ppt
This sequence of tests would definitely identify the bottle contents. 4.29 LiOH is a strong base, HI is a strong acid, and CH3OH is a molecular compound and nonelectrolyte. The strong acid HI will have the greatest concentration of solvated protons. 4.31 (a) A monoprotic acid has one ionizable (acidic) H, whereas a diprotic acid has two. (b) A strong acid is completely ionized in aqueous solution, whereas only a fraction of weak acid molecules are ionized. (c) An acid is an H + donor, and a base is an H + acceptor. 4.33 When each of the strong acids in Table 4.2 dissociates, the anions formed are the same ones that normally form soluble ionic compounds (Table 4.1). The one exception is acetate, CH3COO-, the anion of a weak acid. 4.35 (a) Acid, mixture of ions and molecules (weak electrolyte) (b) none of the above, entirely molecules (nonelectrolyte) (c) salt, entirely ions (strong electrolyte) (d) base, entirely ions (strong electrolyte) 4.37 (a) H2SO3, weak electrolyte (b) C2H5OH, nonelectrolyte (c) NH3, weak electrolyte (d) KClO3, strong electrolyte (e) Cu(NO3)2, strong electrolyte 4.39 (a) 2 HBr(aq) + Ca(OH)2(aq) ¡ CaBr2(aq) + 2 H2O(l); H + (aq) + OH-(aq) ¡ H2O(l) (b) Cu(OH)2(s) + 2 HClO4(aq) ¡ Cu(ClO4)2(aq) + 2 H2O(l); Cu(OH)2(s) + 2 H + (aq) ¡ 2 H2O(l) + Cu2 + (aq) (c) Al(OH)3(s) + 3 HNO3(aq) ¡ Al(NO3)3(aq) + 3 H2O(l); Al(OH)3(s) + 3 H + (aq) ¡ 3 H2O(l) + Al3 + (aq) 4.41 (a) CdS(s) + H2SO4(aq) ¡ CdSO4(aq) + H2S(g); CdS(s) + 2H + (aq) ¡ H2S(g) + Cd2 + (aq) (b) MgCO3(s) + 2 HClO4(aq) ¡ Mg(ClO4)2(aq) + H2O(l) + CO2(g); MgCO3(s) + 2 H + (aq) ¡ H2O(l) + CO2(g) + Mg2 + (aq) 4.43 (a) MgCO3(s) + 2 HCl(aq) ¡ MgCl2(aq) + H2O(l) + CO2(g); MgCO3(s) + 2 H + (aq) ¡ Mg2 + (aq) + H2O(l) + CO2(g); MgO(s) + 2 HCl(aq) ¡ MgCl2(aq) + H2O(l); MgO(s) + 2 H + (aq) ¡ Mg2 + (aq) + H2O(l); Mg(OH)2(s) + 2 HCl(aq) ¡ MgCl2(aq) + 2 H2O(l); Mg(OH)2(s) + 2 H + (aq) ¡ Mg2 + (aq) + 2 H2O(l) (b) Yes. The reaction involving magnesium carbonate, MgCO3(s), produces CO2(g), which appears as bubbles. The other two reactions are calm. (c) If excess HCl(aq) is added in each case, the identity of the ions in the clear product solution is the same. The ions are Mg2+(aq); Cl-(aq); and H+(aq). 4.45 (a) In terms of electron transfer, oxidation is the loss of electrons by a substance and reduction is the gain of electrons (LEO says GER). (b) Relative to oxidation numbers, when a substance is oxidized, its oxidation number increases. When a substance is reduced, its oxidation number decreases. 4.47 Metals in region A are most easily oxidized. Nonmetals in region D are least easily oxidized.
4.49 (a) +4 (b) +4 (c) +7 (d) + 1 (e) 0 (f) -1 4.51 (a) N2 ¡ 2 NH3, N is reduced; 3 H2 ¡ 2 NH3, H is oxidized (b) Fe2 + ¡ Fe, Fe is reduced; Al ¡ Al3 + , Al is oxidized (c) Cl2 ¡ 2 Cl-, Cl is reduced; 2 I- ¡ I2 I is oxidized (d) S2- ¡ SO42- , S is oxidized; H2O2 ¡ H2O, O is reduced 4.53 (a) Mn(s) + H2SO4(aq) ¡ MnSO4(aq) + H2(g); Mn(s) + 2 H + (aq) ¡ Mn2 + (aq) + H2(g) (b) 2 Cr(s) + 6 HBr(aq) ¡ 2 CrBr3(aq) + 3 H2(g); 2 Cr(s) + 6 H + (aq) ¡ 2 Cr 3 + (aq) + 3 H2(g) (c) Sn(s) + 2 HCl(aq) ¡ SnCl 2(aq) + H2(g); Sn(s) + 2 H + (aq) ¡ Sn2 + (aq) + H2(g) (d) 2 Al(s) + 6 HCOOH(aq) ¡ 2 Al(HCOO)3(aq) + 3 H2(g); 2 Al(s) + 6 HCOOH(aq) ¡ 2 Al3 + (aq) + 6 HCOO-(aq) + 3 H2(g) 4.55 (a) Fe(s) + Cu(NO3)2(aq) ¡ Fe(NO3)2(aq) + Cu(s) (b) NR (c) Sn(s) + 2 HBr(aq) ¡ SnBr2(aq) + H2(g) (d) NR (e) 2 Al(s) + 3 CoSO4(aq) ¡ Al 2(SO4)3(aq) + 3 Co(s) 4.57 (a) i. Zn(s) + Cd2 + (aq) ¡ Cd(s) + Zn2 + (aq); ii. Cd(s) + Ni2 + (aq) ¡ Ni(s) + Cd2 + (aq) (b) Cd is between Zn and Ni on the activity series. (c) Place an iron strip in CdCl2(aq). If Cd(s) is deposited, Cd is less active than Fe; if there is no reaction, Cd is more active than Fe. Do the same test with Co if Cd is less active than Fe or with Cr if Cd is more active than Fe. 4.59 (a) Intensive; the ratio of amount of solute to total amount of solution is the same, regardless of how much solution is present. (b) The term 0.50 mol HCl defines an amount ( ' 18 g) of the pure substance HCl. The term 0.50 M HCl is a ratio; it indicates that there is 0.50 mol of HCl solute in 1.0 liter of solution. 4.61 (a) 1.17 M ZnCl2 (b) 0.158 mol HNO3 (c) 54.2 mL of 6.00 M NaOH 4.63 16 g Na + (aq) 4.65 BAC of 0.08 = 0.02 M CH3CH2OH (alcohol) 4.67 (a) 5.21 g KBr (b) 0.06537 M Ca(NO3)2 (c) 10.2 mL of 1.50 M Na3PO4 4.69 (a) 0.15 M K2CrO4 has the highest K + concentration. (b) 30.0 mL of 0.15 M K2CrO4 has more K + ions. 4.71 (a) 0.25 M Na + , 0.25 M NO3(b) 1.3 * 10-2 M Mg 2 + , 1.3 * 10-2 M SO42- (c) 0.0150 M C6H12O6 (d) 0.111 M Na + , 0.111 M Cl- , 0.0292 M NH4 + , 0.0146 M CO324.73 (a) 16.9 mL 14.8 M NH3 (b) 0.296 M NH3 4.75 (a) Add 21.4 g C12H22O11 to a 250-mL volumetric flask, dissolve in a small volume of water, and add water to the mark on the neck of the flask. Agitate thoroughly to ensure total mixing. (b) Thoroughly rinse, clean, and fill a 50-mL buret with the 1.50 M C12H22O11. Dispense 23.3 mL of this solution into a 350-mL volumetric container, add water to the mark, and mix thoroughly. 4.77 1.398 M CH3COOH 4.79 0.227 g KCl 4.81 (a) 38.0 mL of 0.115 M HClO4 (b) 769 mL of 0.128 M HCl (c) 0.408 M AgNO3 (d) 0.275 g KOH 4.83 27 g NaHCO3 4.85 (a) Molar mass of metal hydroxide is 103 g/mol. (b) Rb + 4.87 (a) NiSO4(aq) + 2 KOH(aq) ¡ Ni(OH)2(s) + K2SO4(aq) (b) Ni(OH)2 (c) KOH is the limiting reactant. (d) 0.927 g Ni(OH)2 (e) 0.0667 M Ni2 + (aq), 0.0667 M K + (aq), 0.100 M SO42-(aq) 4.89 91.40% Mg(OH)2 4.92 The precipitate is CdS(s). Na+(aq) and NO3-(aq) remain in solution, along with any excess reactant ions. The net ionic equation is Cd2 + (aq) + S2-(aq) ¡ CdS(s). 4.94 (a, b) Expt. 1: NR; Expt. 2: 2 Ag + (aq) + CrO42-(aq) ¡ Ag 2CrO4(s) red precipitate; Expt. 3: 2 Ca2 + (aq) + CrO42-(aq) ¡ CaCrO4(s) yellow precipitate; Expt. 4: 2 Ag + (aq) + C2O42-(aq) ¡ Ag 2C2O4(s) white precipitate; Expt. 5: Ca2 + (aq) + C2O42-(aq) ¡ CaC2O4(s) white precipitate; Expt. 6: Ag + (aq) + Cl-(aq) ¡ AgCl(s) white precipitate. 4.96 4 NH3(g) + 5 O2(g) ¡ 4 NO(g) + 6 H2O(g). (a) redox reaction (b) N is oxidized, O is reduced. 2 NO(g) + O2(g) ¡ 2 NO2(g). (a) redox reaction (b) N is oxidized, O is reduced. 3 NO2(g) + H2O(l) ¡ HNO3(aq) + NO(g). (a) redox reaction (b) N is both oxidized and reduced. 4.99 1.42 M KBr 4.100 (a) 2.2 * 10-9 M Na + (b) 1.3 * 1012 Na + ions 4.103 (a) 1.718 M Sr(OH)2 (b) 2 HNO3(aq) + Sr(OH)2(s) ¡ Sr(NO3)2(aq) + 2 H2O(l) (c) 2.61 M HNO3 4.106 (a) The molarmass of the acid is 136 g>mol. (b) The molecular formula is C8H8O2. 4.109 (a) Mg(OH)2(s) + 2 HNO3(aq) ¡ Mg (NO3)2(aq) + 2 H2O(l) (b) HNO3 is the limiting reactant. (c) 0.0923 mol Mg(OH)2, 0 mol HNO3, and 0.00250 mol Mg(NO3)2 are present. 4.112 1.766% Cl- by mass 4.114 1.5 * 10-5 g Na3AsO4 in 1.00 L H2O
Answers to Selected Exercises
CHAPTER 5 5.1 (a) As the book falls, potential energy decreases and kinetic energy increases. (b) 71 J, assuming no transfer of energy as heat (c) A heavier book falling from the same shelf has greater kinetic energy when it hits the floor. 5.5 (a) No. The distance traveled to the top of a mountain depends on the path taken by the hiker. Distance is a path function, not a state function. (b) Yes. Change in elevation depends only on the location of the base camp and the height of the mountain, not on the path to the top. Change in elevation is a state function, not a path function. 5.8 (a) The sign of w is ( +). (b) The internal energy of the system increases during the change; the sign of ΔE is ( +). 5.11 (a) ¢HA = ¢HB + ¢HC. The diagram and equation both show that the net enthalpy change for a process is independent of path, that ¢H is a state function. (b) ¢HZ = ¢HX + ¢HY. (c) Hess’s law states that the enthalpy change for net reaction Z is the sum of the enthalpy changes for steps X and Y, regardless of whether the reaction actually occurs via this path. The diagrams are a visual statement of Hess’s law. 5.13 An object can possess energy by virtue of its motion or position. Kinetic energy depends on the mass of the object and its velocity. Potential energy depends on the position of the object relative to the body with which it interacts. 5.15 (a) 1.9 * 105 J (b) 4.6 * 104 cal (c) As the automobile brakes to a stop, its speed (and hence its kinetic energy) drops to zero. The kinetic energy of the automobile is primarily transferred to friction between brakes and wheels and somewhat to deformation of the tire and friction between the tire and road. 5.17 1 Btu = 1054 J 5.19 (a) The system is the well-defined part of the universe whose energy changes are being studied. (b) A closed system can exchange heat but not mass with its surroundings. (c) Any part of the universe not part of the system is called the surroundings. 5.21 (a) Work is a force applied over a distance. (b) The amount of work done is the magnitude of the force times the distance over which it is applied. w = F * d. 5.23 (a) Gravity; work is done because the force of gravity is opposed and the pencil is lifted. (b) Mechanical force; work is done because the force of the coiled spring is opposed as the spring is compressed over a distance. 5.25 (a) In any chemical or physical change, energy can be neither created nor destroyed; energy is conserved. (b) The internal energy (E) of a system is the sum of all the kinetic and potential energies of the system components. (c) Internal energy of a closed system increases when work is done on the system and when heat is transferred to the system. 5.27 (a) ¢E = - 0.077 kJ, endothermic (b) ¢E = - 22.1 kJ, exothermic (c) ¢E = 7.25 kJ, endothermic 5.29 (a) Since no work is done by the system in case (2), the gas will absorb most of the energy as heat; the case (2) gas will have the higher temperature. (b) In case (2) w = 0 and q = 100 J. In case (1) energy will be used to do work on the surroundings (- w), but some will be absorbed as heat (+ q). (c) ¢E is greater for case (2) because the entire 100 J increases the internal energy of the system rather than a part of the energy doing work on the surroundings. 5.31 (a) A state function is a property that depends only on the physical state (pressure, temperature, etc.) of the system, not on the route used to get to the current state. (b) Internal energy is a state function; heat is not a state function. (c) Volume is a state function. The volume of a system depends only on conditions (pressure, temperature, amount of substance), not the route or method used to establish that volume. 5.33 (a) ¢H is usually easier to measure than ¢E because at constant pressure, ¢H = qp. The heat flow associated with a process at constant pressure can easily be measured as a change in temperature, while measuring ¢E requires a means to measure both q and w. (b) H is a static quantity that depends only on the specific conditions of the system. q is an energy change that, in the general case, does depend on how the change occurs. We can equate change in enthalpy, ¢H, with heat, qp, only for the specific conditions of constant pressure and exclusively P-V work. (c) The process is endothermic. 5.35 At constant pressure, ¢E = ¢H - P ¢V. The values of either P and ¢V or T and ¢n must be known to calculate ¢E from ¢H. 5.37 ΔE = 1.47 kJ; ΔH = 0.824 kJ 5.39 (a) C2H5OH(l) + 3 O2(g) ¡ 3 H2O + 2 CO2(g), ¢H = - 1235 kJ
A-5
(b) C2H5OH(l) 3 O2(g)
H 1235 kJ
3 H2O(g) 2 CO2(g)
5.41 (a) ¢H = - 142.3 kJ>mol O3(g) (b) 2 O3(g) has the higher enthalpy. 5.43 (a) Exothermic (b) -87.9 kJ heat transferred (c) 15.7 g MgO produced (d) 602 kJ heat absorbed 5.45 (a) - 29.5 kJ (b) - 4.11 kJ (c) 60.6 J 5.47 (a) ¢H = 726.5 kJ (b) ¢H = - 1453 kJ (c) The exothermic forward reaction is more likely to be thermodynamically favored. (d) Vaporization is endothermic. If the product were H2O(g), the reaction would be more endothermic and would have a less negative ¢H. 5.49 (a) J>mol-°C or J>mol-K (b) J>g-°C or J>g-K (c) To calculate heat capacity from specific heat, the mass of the particular piece of copper pipe must be known. 5.51 (a) 4.184 J>g-K (b) 75.40 J>mol-°C (c) 774 J>°C (d) 904 kJ 5.53 (a) 2.66 * 103 J (b) It will require more heat to increase the temperature of one mole of octane, C8H18(l), by a certain amount than to increase the temperature of one mole of water, H2O(l), by the same amount. 5.55 ¢H = - 44.4 kJ>mol NaOH 5.57 ¢Hrxn = - 25.5 kJ>g C6H4O2 or -2.75 * 103 kJ>mol C6H4O2 5.59 (a) Heat capacity of the complete calorimeter = 14.4 kJ>°C (b) 7.56 °C 5.61 Hess’s law is a consequence of the fact that enthalpy is a state function. Since ¢H is independent of path, we can describe a process by any series of steps that adds up to the overall process. ¢H for the process is the sum of ¢H values for the steps. 5.63 ¢H = - 1300.0 kJ 5.65 ¢H = - 2.49 * 103 kJ 5.67 (a) Standard conditions for enthalpy changes are P = 1 atm and some common temperature, usually 298 K. (b) Enthalpy of formation is the enthalpy change that occurs when a compound is formed from its component elements. (c) Standard enthalpy of formation ¢Hf° is the enthalpy change that accompanies formation of one mole of a substance from elements in their standard states. 5.69 (a) 12 N2(g) + O2(g) ¡ NO2(g), ¢Hf° = 33.84 kJ (b) S(s) + 3>2 O2(g) ¡ SO3(g), ¢H f° = - 395.2 kJ (c) Na(s) + 1 2 Br2(l) ¡ NaBr(s), ¢H f° = - 361.4 kJ (d) Pb(s) + N2(g) + ° = 3 O2(g) ¡ Pb(NO3)2(s) , ¢H f° = - 451.9 kJ 5.71 ¢Hrxn ° = - 196.6 kJ (b) ¢Hrxn ° = 37.1 kJ - 847.6 kJ 5.73 (a) ¢Hrxn ° = - 976.94 kJ (d) ¢H rxn ° = - 68.3 kJ 5.75 ¢H f° = - 248 kJ (c) ¢H rxn 5.77 (a) C8H18 (l) + 25 2 O2(g) ¡ 8 CO2(g) + 9 H2O(g), (b) ¢H = -5064.9 kJ 8 C(s, graphite) + 9 H2(g) ¡ C8H18(l) (c) ¢H f° = -259.5 kJ 5.79 (a) C2H5OH(l) + 3 O2(g) ¡ 2 CO2(g) + ° = - 1234.8 kJ (c) 2.11 * 104 kJ>L heat produced 3 H2O(g) (b) ¢H rxn (d) 0.071284 g CO2>kJ heat emitted 5.81 (a) Fuel value is the amount of energy produced when 1 g of a substance (fuel) is combusted. (b) 5 g of fat (c) These products of metabolism are expelled as waste via the alimentary tract, H2O(l) primarily in urine and feces, and CO2(g) as gas. 5.83 108 or 1 * 102 Cal>serving (b) Sodium does not contribute to the calorie content of the food because it is not metabolized by the body. 5.85 59.7 Cal 5.87 (a) ¢Hcomb = - 1850 kJ>mol C3H4, -1926 kJ>mol C3H6, -2044 kJ>mol C3H8 (b) ¢Hcomb = - 4.616 * 104 kJ>kg C3H4, - 4.578 * 104 kJ>kg C3H6, -4.635 * 104 kJ>kg C3H8 (c) These three substances yield nearly identical quantities of heat per unit mass, but propane is marginally higher than the other two. 5.89 1 * 1012 kg C6H12O6/yr 5.91 (a) 469.4 m>s (b) 5.124 * 10-21 J (c) 3.086 kJ>mol 5.93 The spontaneous air bag reaction is probably exothermic, with - ¢H and thus - q. When the bag inflates, work is done by the system, so the sign of w is also negative. 5.97 ¢H = 38.95 kJ; ¢E = 36.48 kJ ° = -353.0 kJ (b) 1.2 g Mg needed 5.106 (a) ¢H ° = 5.102 (a) ¢H rxn -631.3 kJ (b) 3 mol of acetylene gas has greater enthalpy. (c) Fuel values are 50 kJ>g C2H2(g), 42 kJ>g C6H6(l). 5.109 If all work is used to increase the man’s potential energy, the stair climbing uses 58 Cal and will not compensate for the extra order of 245 Cal fries. (More than 58 Cal will be required to climb the stairs because some energy is used to move limbs and some will be lost as heat.)
A-6
Answers to Selected Exercises
5.112 (a) 1.479 * 10-18 J>molecule (b) 1 * 10-15 J>photon. The X-ray has approximately 1000 times more energy than is produced by the combustion of 1 molecule of CH4(g). 5.114 (a) ¢H° for neutralization of the acids is HNO3, -55.8 kJ; HCl, -56.1 kJ; NH4+ , -4.1 kJ. (b) H + (aq) + OH-(aq) ¡ H2O(l) is the net ionic equation for the first two reactions. NH4+ (aq) + OH-(aq) ¡ NH3(aq) + H2O(l) (c) The ¢H° values for the first two reactions are nearly identical, -55.8 kJ and -56.1 kJ. Since spectator ions do not change during a reaction and these two reactions have the same net ionic equation, it is not surprising that they have the same ¢H°. (d) Strong acids are more likely than weak acids to donate H + . Neutralization of the two strong acids is energetically favorable, while the third reaction is barely so. NH4 + is likely a weak acid. 5.116 (a) ¢H° = -65.7 kJ (b) ¢H° for the complete molecular equation will be the same as ¢H° for the net ionic equation. Since the overall enthalpy change is the enthalpy of products minus the enthalpy of reactants, the contributions of spectator ions cancel. (c) ¢Hf° for AgNO3(aq) is -100.4 kJ>mol.
know the position and momentum (a quantity related to energy) of an electron. The Bohr model states that electrons move about the nucleus in precisely circular orbits of known radius and energy. This violates the uncertainty principle. (b) De Broglie stated that electrons demonstrate the properties of both particles and waves and that each moving particle has a wave associated with it. A wave function is the mathematical description of the matter wave of an electron. (c) Although we cannot predict the exact location of an electron in an allowed energy state, we can determine the probability of finding an electron at a particular position. This statistical knowledge of electron location is the probability density and is a function of C 2, the square of the wave function C. 6.51 (a) n = 4, l = 3, 2, 1, 0 (b) l = 2, ml = -2, -1, 0, 1, 2 (c) ml = 2, l Ú 2 or l = 2, 3 or 4 6.53 (a) 3p: n = 3, l = 1 (b) 2s: n = 2, l = 0 (c) 4f: n = 4, l = 3 (d) 5d: n = 5, l = 2 6.55 (a) impossible, 1p (b) possible (c) possible (d) impossible, 2d 6.57 (a)
z
(b)
CHAPTER 6 6.2 (a) 0.1 m or 10 cm (b) No. Visible radiation has wavelengths much shorter than 0.1 m. (c) Energy and wavelength are inversely proportional. Photons of the longer 0.1-m radiation have less energy than visible photons. (d) Radiation with l = 0.1 m is in the low-energy portion of the microwave region. The appliance is probably a microwave oven. 6.5 (a) Increase (b) decrease (c) the light from the hydrogen discharge tube is a line spectrum, so not all visible wavelengths will be in our “hydrogen discharge rainbow.” Starting on the inside, the rainbow will be violet, then blue and blue-green. After a gap, the final band will be red. 6.8 (a) 1 (b) p (c) For the n = 4 shell, the lobes in the contour representation would extend farther along the y-axis. 6.11 (a) Meters (b) 1>second (c) meters>second 6.13 (a) True (b) False. Ultraviolet light has shorter wavelengths than visible light. (c) False. X-rays travel at the same speed as microwaves. (d) False. Electromagnetic radiation and sound waves travel at different speeds. 6.15 Wavelength of X-rays 6 ultraviolet 6 green light 6 red light 6 infrared 6 radio waves 6.17 (a) 3.0 * 1013 s-1 (b) 5.45 * 10-7 m = 545 nm (c) The radiation in (b) is visible; the radiation in (a) is not. (d) 1.50 * 104 m 6.19 5.64 * 1014 s-1; green. 6.21 Quantization means that energy changes can happen only in certain allowed increments. If the human growth quantum is one foot, growth occurs instantaneously in one-foot increments. The child experiences growth spurts of one foot; her height can change only by one-foot increments. 6.23 (a) 4.47 * 10-21 J (b) 6.17 * 10-19 J (c) 69.2 nm 6.25 (a) l = 3.3 mm, E = 6.0 * 10-20 J; l = 0.154 nm, E = 1.29 * 10-15 J (b) The 3.3-mm photon is in the infrared region and the 0.154-nm photon is in the X-ray region; the X-ray photon has the greater energy. 6.27 (a) 6.11 * 10-19 J>photon (b) 368 kJ>mol (c) 1.64 * 1015 photons (d) 368 kJ>mol 6.29 (a) The ' 1 * 10 - 6 m radiation is in the infrared portion of the spectrum. (b) 8.1 * 1016 photons>s 6.31 (a) Emin = 7.22 * 10-19 J (b) l = 275 nm (c) E120 = 1.66 * 10-18 J. The excess energy of the 120-nm photon is converted into the kinetic energy of the emitted electron. Ek = 9.3 * 10-19 J>electron. 6.33 When applied to atoms, the notion of quantized energies means that only certain values of ¢E are allowed. These are represented by the lines in the emission spectra of excited atoms. 6.35 (a) Emitted (b) absorbed (c) emitted 6.37 (a) E2 = -5.45 * 10-19 J; E6 = -0.606 * 10-19 J; ¢E = 4.84 * 10-19 J; l = 410 nm (b) visible, violet 6.39 (a) Only lines with nf = 2 represent ¢E values and wavelengths that lie in the visible portion of the spectrum. Lines with nf = 1 have shorter wavelengths and lines with nf 7 2 have longer wavelengths than visible radiation. (b) ni = 3, nf = 2; l = 6.56 * 10-7 m; this is the red line at 656 nm. ni = 4, nf = 2; l = 4.86 * 10-7 m; this is the blue-green line at 486 nm. ni = 5, nf = 2; l = 4.34 * 10-7 m; this is the blue-violet line at 434 nm. 6.41 (a) Ultraviolet region (b) ni = 6, nf = 1 6.43 (a) l = 5.6 * 10-37 m (b) l = 2.65 * 10-34 m (c) l = 2.3 * 10-13 m (d) l = 1.51 * 10-11 m 6.45 4.14 * 103 m>s 6.47 (a) ¢x Ú 4 * 10-27 m (b) ¢x Ú 3 * 10-10 m 6.49 (a) The uncertainty principle states that there is a limit to how precisely we can simultaneously
z
(c)
y x
s
y x
pz
x
y z
d xy
6.59 (a) The hydrogen atom 1s and 2s orbitals have the same overall spherical shape, but the 2s orbital has a larger radial extension and one more node than the 1s orbital. (b) A single 2p orbital is directional in that its electron density is concentrated along one of the three Cartesian axes of the atom. The dx2-y2 orbital has electron density along both the x- and y-axes, while the px orbital has density only along the x-axis. (c) The average distance of an electron from the nucleus in a 3s orbital is greater than for an electron in a 2s orbital. (d) 1s 6 2p 6 3d 6 4f 6 6s 6.61 (a) In the hydrogen atom, orbitals with the same principal quantum number, n, have the same energy. (b) In a many-electron atom, for a given n value, orbital energy increases with increasing l value: s 6 p 6 d 6 f. 6.63 (a) There are two main pieces of experimental evidence for electron “spin.” The Stern-Gerlach experiment shows that atoms with a single unpaired electron interact differently with an inhomogeneous magnetic field. Examination of the fine details of emission line spectra of multi-electron atoms reveals that each line is really a close pair of lines. Both observations can be rationalized if electrons have the property of spin. (c)
(b) 2s
2s
1s
1s
6.65 (a) 6 (b) 10 (c) 2 (d) 14 6.67 (a) “Valence electrons” are those involved in chemical bonding. They are part or all of the outer-shell electrons listed after the core. (b) “Core electrons” are inner-shell electrons that have the electron configuration of the nearest noble-gas element. (c) Each box represents an orbital. (d) Each half-arrow in an orbital diagram represents an electron. The direction of the half-arrow represents electron spin. 6.69 (a) Cs, [Xe]6s1 (b) Ni, [Ar]4s23d8 (c) Se, [Ar]4s23d104p4 (d) Cd, [Kr]5s24d10 (e) U, [Rn]5f 36d17s2 (f) Pb, [Xe]6s24f 145d106p2 6.71 (a) Be, 0 unpaired electrons (b) O, 2 unpaired electrons (c) Cr, 6 unpaired electrons (d) Te, 2 unpaired electrons 6.73 (a) The fifth electron would fill the 2p subshell before the 3s. (b) Either the core is [He], or the outer electron configuration should be 3s23p3. (c) The 3p subshell would fill before the 3d. 6.75 (a) lA = 3.6 * 10-8 m, lB = 8.0 * 10-8 m (b) nA = 8.4 * 1015 s-1, nB = 3.7 * 1015 s-1 (c) A, ultraviolet; B, ultraviolet 6.78 66.7 min 6.82 1.6 * 107 photons/s, 5.1 * 10-12 J/s 6.85 (a) The Paschen series lies in the infrared. (b) ni = 4, l = 1.87 * 10-6 m; ni = 5, l = 1.28 * 10-6 m; ni = 6, l = 1.09 * 10-6 m 6.90 (a) l (b) n and l (c) ms (d) ml 6.92 (a) The nodal plane of the pz orbital is the xy-plane. (b) The two nodal planes of the dxy orbital are the ones where
Answers to Selected Exercises x = 0 and y = 0. These are the yz- and xz-planes. (c) The two nodal planes of the dx2-y2 orbital are the ones that bisect the x- and y-axes and contain the z-axis. 6.94 If ms had three allowed values instead of two, each orbital would hold three electrons instead of two. Assuming that there is no change in the n, l, and ml values, the number of elements in each of the first four rows would be 1st row, 3 elements; 2nd row, 12 elements; 3rd row, 12 elements; 4th row, 27 elements 6.97 (a) 1.7 * 1028 photons (b) 34 s 6.101 (a) Bohr’s theory was based on the Rutherford nuclear model of the atom: a dense positive charge at the center and a diffuse negative charge surrounding it. Bohr’s theory then specified the nature of the diffuse negative charge. The prevailing theory before the nuclear model was Thomson’s plum pudding model: discrete electrons scattered about a diffuse positive charge cloud. Bohr’s theory could not have been based on the Thomson model of the atom. (b) De Broglie’s hypothesis is that electrons exhibit both particle and wave properties. Thomson’s conclusion that electrons have mass is a particle property, while the nature of cathode rays is a wave property. De Broglie’s hypothesis actually rationalizes these two seemingly contradictory observations about the properties of electrons.
CHAPTER 7 7.3 (a) The bonding atomic radius of A, rA, is d1>2; rx = d2 - (d1>2). (b) The length of the X ¬ X bond is 2rx or 2d2 - d1. 7.6 (a) X + 2F2 : XF4 (b) X in the diagram has about the same bonding radius as F, so it is likely to be a nonmetal. 7.7 The number of columns in the various blocks of the periodic chart corresponds to the maximum number of electrons that can occupy the various kinds of atomic orbitals: 2 columns on the left for 2 electrons in s orbitals, 10 columns in the transition metals for 10 electrons in d orbitals, 6 columns on the right for 6 electrons in p orbitals, 14-member rows below for 14 electrons in f orbitals. The order of blocks corresponds to the filling order of atomic orbitals, and the row number corresponds to the principal quantum number of the valence electrons of elements in that row, ns, np, (n - 1)d, (n - 2)f. 7.9 In general, elements are discovered according to their ease of isolation in elemental form. 7.11 (a) Effective nuclear charge, Zeff, is a representation of the average electrical field experienced by a single electron. It is the average environment created by the nucleus and the other electrons in the molecule, expressed as a net positive charge at the nucleus. (b) Going from left to right across a period, effective nuclear charge increases. 7.13 (a) For both Na and K, Zeff = 1. (b) For both Na and K, Zeff = 2.2. (c) Slater’s rules give values closer to the detailed calculations: Na, 2.51; K, 3.49. (d) Both approximations give the same value of Zeff for Na and K; neither accounts for the gradual increase in Zeff moving down a group. (e) Following the trend from detailed calculations, we predict a Zeff value of approximately 4.5. 7.15 The n = 3 electrons in Kr experience a greater effective nuclear charge and thus have a greater probability of being closer to the nucleus. 7.17 (a) Atomic radii are determined by measuring distances between atoms in various situations. (b) Bonding radii are calculated from the internuclear separation of two atoms joined by a covalent chemical bond. Nonbonding radii are calculated from the internuclear separation between two gaseous atoms that collide and move apart but do not bond. (c) For a given element, the nonbonding radius is always larger than the bonding radius. (d) If a free atom reacts to become part of a covalent molecule, its radius changes from nonbonding to bonding and the atom gets smaller. 7.19 (a) 1.37 Å (b) The distance between W atoms will decrease. 7.21 From the sum of the atomic radii, As ¬ I = 2.52 Å. This is very close to the experimental value of 2.55 Å. 7.23 (a) Decrease (b) increase (c) O 6 Si 6 Ge 6 I 7.25 (a) Cs 7 K 7 Li (b) Pb 7 Sn 7 Si (c) N 7 O 7 F 7.27 (a) False (b) true (c) false 7.29 The red sphere is a metal; its size decreases on reaction, characteristic of the change in radius when a metal atom forms a cation. The blue sphere is a nonmetal; its size increases on reaction, characteristic of the change in radius when a nonmetal atom forms an anion. 7.31 (a) An isoelectronic series is a group of atoms or ions that have the same number of electrons and the same electron configuration. (b) Ga3 + : Ar; Zr4 + : Kr; Mn7 + : Ar; I- : Xe; Pb2 + : Hg 7.33 (a) Ar (b) Ar (c) There is no neutral atom isoelectronic with Fe2 + . Because transition metals fill the s subshell first but
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also lose s electrons first when they form ions, many transition metal ions do not have isolectronic neutral atoms. (d) No isoelectronic neutral atom; same reason as part (c). (e) No isoelectronic neutral atom; same reason as part (c). 7.35 (a) K+ is smaller. (b) Cl-, Zeff = 7; K + , Zeff = 9 (c) Cl-: Zeff = 5.75; K + , Zeff = 7.75 (d) For isoelectronic ions, as nuclear charge (Z) increases, effective nuclear charge (Zeff) increases and ionic radius decreases. 7.37 (a) Se 6 Se2- 6 Te2(b) Co3 + 6 Fe3 + 6 Fe2 + (c) Ti4 + 6 Sc3 + 6 Ca (d) Be 2 + 6 Na + 6 Ne 7.39 Al(g) ¡ Al + (g) + le - ; Al + (g) ¡ Al2 +(g) + le - ; Al2 + (g) ¡ Al3 + (g) + le- . The process for the first ionization energy requires the least amount of energy. 7.41 (a) False. Ionization energies are always positive quantities. (b) False. F has a greater first ionization energy than O. (c) True. 7.43 (a) The smaller the atom, the larger its first ionization energy. (b) Of the nonradioactive elements, He has the largest and Cs the smallest first ionization energy. 7.45 (a) Cl (b) Ca (c) K (d) Ge (e) Sn 7.47 (a) Fe2 + , 3Ar43d6 (b) Hg 2 + , 3Xe44f 145d10 (c) Mn2 + , 3Ar43d5 (d) Pt2 + , 3Xe44f 145d8 (e) P3- , 3Ne43s23p6 7.49 Ni2 + , 3Ar43d8; Pd2 + , 3Kr44d8; Pt2 + , 3Xe44f 145d8 7.51 (a) Positive, endothermic, values for ionization energy and electron affinity mean that energy is required to either remove or add electrons. Valence electrons in Ar experience the largest Zeff of any element in the third row, resulting in a large, positive ionization energy. When an electron is added to Ar, the n = 3 electrons become core electrons that screen the extra electron so effectively that Ar- has a higher energy than an Ar atom and a free electron. This results in a large positive electron affinity. (b) kJ/mol 7.53 Electron affinity of Br: Br(g) + l e ¡ Br (g); 2 10 5 2 10 6 3Ar44s 3d 4p ¡ 3Ar44s 3d 4p ; electron affinity of Kr: Kr(g) + 1 e- ¡ Kr-(g); 3Ar44s23d104p6 ¡ 3Ar44s23d104p65s1. Br- adopts the stable electron configuration of Kr; the added electron experiences essentially the same Zeff and stabilization as the other valence electrons and electron affinity is negative. In Kr- ion, the added electron occupies the higher energy 5s orbital. A 5s electron is farther from the nucleus, effectively shielded by the spherical Kr core and not stabilized by the nucleus; electron affinity is positive. 7.55 (a) Ionization energy (I1) of Ne: Ne(g) ¡ Ne + (g) + 1 e-; 3He42s22p6 ¡ 3He42s22p5 ; electron affinity (E1) of F: F(g) + 1 e- ¡ F-(g); 3He42s22p5 ¡ 3He42s22p6. (b) I1 of Ne is positive; E1 of F is negative. (c) One process is apparently the reverse of the other, with one important difference. Ne has a greater Z and Zeff, so we expect I1 for Ne to be somewhat greater in magnitude and opposite in sign to E1 for F. 7.57 The smaller the first ionization energy of an element, the greater the metallic character of that element. 7.59 Agree. When forming ions, all metals form cations. The only nonmetallic element that forms cations is the metalloid Sb, which is likely to have significant metallic character. 7.61 Ionic: SnO2, Al2O3, Li2O, Fe2O3; molecular: CO2, H2O. Ionic compounds are formed by combining a metal and a nonmetal; molecular compounds are formed by two or more nonmetals. 7.63 (a) An acidic oxide dissolved in water produces an acidic solution; a basic oxide dissolved in water produces a basic solution. (b) Oxides of nonmetals, such as SO3, are acidic; oxides of metals, such as CaO, are basic. 7.65 (a) Dichlorineheptoxide (b) 2 Cl2(g) + 7 O2(g) ¡ 2 Cl2O7(l) (c) While most nonmetal oxides we have seen, such as CO2 or SO2, are gases, a boiling point of 81 °C is expected for a large molecule like Cl2O7. (d) Cl2O7 is an acidic oxide, so it will be more reactive to base, OH-. (e) The oxidation state of Cl in Cl2O7 is +7; the corresponding electron configuration for Cl is 3He42s22p6 or [Ne]. 7.67 (a) BaO(s) + H2O(l) ¡ Ba(OH)2(aq) (b) FeO(s) + 2 HClO4(aq) ¡ Fe(ClO4)2(aq) + H2O(l) (c) SO3(g) + (d) H2O(l) ¡ H2SO4(aq) CO2(g) + 2 NaOH(aq) ¡ Na2CO3(aq) + H2O(l) 7.69 Yes, the reactivity of a metal correlates with its first ionization energy. Since metals lose electrons when they form ions, the less energy required for this process, the more reactive the metal. However, we usually observe reactivity of metals in the solid state and ionization energy is a gas phase property, so there are differences between the two properties. 7.71 (a) Ca is more reactive because it has a lower ionization energy than Mg. (b) K is more reactive because it has a lower ionization energy than Ca. 7.73 (a) 2 K(s) + Cl2(g) ¡ 2 KCl(s) (b) SrO(s) + H2O(l) ¡ Sr(OH)2(aq) (c) 4 Li(s) + O2(g) ¡ 2 Li2O(s) (d) 2 Na(s) + S(l) ¡ Na2S(s) 7.75 (a) Both classes of reaction are redox
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Answers to Selected Exercises
reactions where either hydrogen or the halogen gains electrons and is reduced. The product is an ionic solid, where either hydride ion, H- , or a halide ion, X-, is the anion. (b) Ca(s) + F2(g) ¡ CaF2(s); Ca(s) + H2(g) ¡ CaH2(s). Both products are ionic solids containing Ca2 + and the corresponding anion in a 1:2 ratio. 7.77 (a) Br, 3Ar44s24p5; Cl, 3Ne43s23p5 (b) Br and Cl are in the same group, and both adopt a 1- ionic charge. (c) The ionization energy of Br is smaller than that of Cl, because the 4p valence electrons in Br are farther from to the nucleus and less tightly held than the 3p electrons of Cl. (d) Both react slowly with water to form HX + HOX. (e) The electron affinity of Br is less negative than that of Cl, because the electron added to the 4p orbital in Br is farther from the nucleus and less tightly held than the electron added to the 3p orbital of Cl. (f) The atomic radius of Br is larger than that of Cl, because the 4p valence electrons in Br are farther from the nucleus and less tightly held than the 3p electrons of Cl. 7.79 (a) The term inert was dropped because it no longer described all the Group 8A elements. (b) In the 1960s, scientists discovered that Xe would react with substances having a strong tendency to remove electrons, such as F2. Thus, Xe could not be categorized as an “inert” gas. (c) The group is now called the noble gases. 7.81 (a) 2 O3(g) ¡ 3 O2(g) (b) Xe(g) + F2(g) ¡ XeF2(g); Xe(g) + 2 F2(g) ¡ XeF4(s); Xe(g) + 3 F2(g) ¡ XeF6(s) (c) S(s) + (d) 2 F2(g) + 2 H2O(l) ¡ 4 HF(aq) + O2(g) 7.83 Up to Z = 82, there are three instances where atomic weights are reversed relative to atomic numbers: Ar and K; Co and Ni; Te and I. In each case the most abundant isotope of the element with the larger atomic number has one more proton but fewer neutrons than the element with the smaller atomic number. The smaller number of neutrons causes the element with the larger Z to have a smaller than expected atomic weight. 7.85 (a) 5 + (b) 4.8 + (c) Shielding is greater for 3p electrons, owing to penetration by 3s electrons, so Zeff for 3p electrons is less than that for 3s electrons. (d) The first electron lost is a 3p electron because it has a smaller Zeff and experiences less attraction for the nucleus than a 3s electron does. 7.88 (a) The estimated distances are a bit longer than the measured distances. This probably shows a systematic bias in either the estimated radii or in the method of obtaining the measured values. (b) The principal quantum number of the outer electrons and thus the average distance of these electrons from the nucleus increase from P(n = 3) to As(n = 4) to Sb(n = 5). This causes the systematic increase in M – H distance. 7.92 (a) 2 Sr(s) + O2(g) ¡ 2 SrO(s) (b) Based on ionic radii, the length of the side of the cube is 5.16 Å. (c) There are four SrO units in the cube.
ionization energy, 805 kJ/mol; electron affinity, -235 kJ/mol; atomic size, 1.65 Å; common oxidation state, -1 7.106 (a) Li, 3He42s1; Zeff L 1+ (b) I1 L 5.45 * 10-19 J>mol L 328 kJ>mol (c) The estimated value of 328 kJ>mol is less than the Table 7.4 value of 520 kJ>mol. Our estimate for Zeff was a lower limit; the [He] core electrons do not perfectly shield the 2s electron from the nuclear charge. (d) Based on the experimental ionization energy, Zeff = 1.26. This value is greater than the estimate from part (a) but agrees well with the “Slater” value of 1.3 and is consistent with the explanation in part (c). 7.108 (a) 9.8902 Å. (b) For Hg, the first ionization energy is 1007 kJ/mol, while the XPS energy of the 4f electron is 10,100 kJ/mol. The energy required to remove a 4f core electron is 10 times the energy required to remove a 6s valence electron. For O, the first ionization energy is 1314 kJ/mol, while the XPS energy of a 1s electron is 51,200 kJ/mol. The energy required to remove a 1s core electron is 40 times that required to remove a 2p valence electron. (c) Hg 2 + , 3Xe44f145d10; valence electrons are 5d. O2- , 3He42s22p6 or [Ne]; valence electrons are 2p (d) Hg2 + 5d, Zeff = 18.85; Hg2 + 4f, Zeff = 43.85; O2-4f, Zeff = 3.85. 7.110 (a) Mg3N2 (b) Mg3N2(s) + 3 H2O(l) ¡ 3 MgO(s) + 2 NH3(g); the driving force is the production of NH3(g). (c) 17% Mg3N2 (d) 3 Mg(s) + 2 NH3(g) ¡ Mg3N2(s) + 3 H2(g). NH3 is the limiting reactant and 0.46 g H2 is formed. ° = - 368.70 kJ (e) ¢Hrxn
7.95 (a) O, [He]2s22p4
8.17 (a) AlF3 (b) K2S (c) Y2O3 (d) Mg 3N2 8.19 (a) Sr 2 + , 3Ar44s23d104p6 = 3Kr4, noble-gas configuration (b) Ti2 + , 3Ar43d2 (c) Se 2- , 3Ar44s23d104p6 = 3Kr4, noble-gas configuration (d) Ni 2 + , 3Ar43d8 (e) Br - , 3Ar44s23d104p6 = 3Kr4, noble-gas configuration (f) Mn3 + , 3Ar43d4 8.21 (a) Lattice energy is the energy required to totally separate one mole of solid ionic compound into its gaseous ions. (b) The magnitude of the lattice energy depends on the magnitudes of the charges of the two ions, their radii, and the arrangement of ions in the lattice. 8.23 KF, 808 kJ>mol; CaO, 3414 kJ>mol; ScN, 7547 kJ>mol. The interionic distances in the three compounds are similar. For compounds with similar ionic separations, the lattice energies should be related as the product of the charges of the ions. The lattice energies above are approximately related as 1 : 4 : 9. Slight variations are due to the small differences in ionic separations. 8.25 Since the ionic charges are the same in the two compounds, the K ¬ Br and Cs ¬ Cl separations must be approximately equal. 8.27 The large attractive energy between oppositely charged Ca2 + and O2- more than compensates for the energy required to form Ca2 + and O2- from the neutral atoms. 8.29 The lattice energy of RbCl(s) is +692 kJ>mol. This value is smaller than the lattice energy for NaCl because Rb + has a larger ionic radius than Na + and therefore cannot approach Cl- as closely as Na + can. 8.31 (a) A covalent bond is the bond formed when two atoms share one or more pairs of electrons. (b) Any simple compound whose component atoms are nonmetals, such as H2, SO2, and CCl 4, are molecular and have covalent bonds between atoms. (c) Covalent, because it is a gas at room temperature and below.
2s
2p
O2, [He]2s22p6 [Ne] 2s 2p (b) O3- , 3Ne43s1. The third electron would be added to the 3s orbital, which is farther from the nucleus and more strongly shielded by the [Ne] core. The overall attraction of this 3s electron for the oxygen nucleus is not large enough for O3- to be a stable particle. 7.98 (a) For both H and the alkali metals, the added electron will complete an ns subshell, so shielding and repulsion effects will be similar. For the halogens, the electron is added to an np subshell, so the energy change is likely to be quite different. (b) True. The electron configuration of H is 1s1. The single 1s electron experiences no repulsion from other electrons and feels the full unshielded nuclear charge. The outer electrons of all other elements that form compounds are shielded by a spherical inner core of electrons and are less strongly attracted to the nucleus, resulting in larger bonding atomic radii. (c) Both H and the halogens have large ionization energies. The relatively large effective nuclear charge experienced by np electrons of the halogens is similar to the unshielded nuclear charge experienced by the H 1s electron. For the alkali metals, the ns electron being removed is effectively shielded by the core electrons, so ionization energies are low. (d) ionization energy of hydride, H-(g) ¡ H(g) + 1 e - (e) electron affinity of hydrogen, H(g) + 1 e - ¡ H-(g). The value for the ionization energy of hydride is equal in magnitude but opposite in sign to the electron affinity 2 14 10 5 of hydrogen. 7.103 Electron configuration, 3Rn47s 5f 6d 7p ; first
CHAPTER 8 8.1 (a) Group 4A or 14 (b) Group 2A or 2 (c) Group 5A or 15 8.4 (a) Ru (b) 3Kr45s24d6. 8.7 (a) Moving from left to right along the molecule, the first C needs 2 H atoms, the second needs 1, the third needs none, and the fourth needs 1. (b) In order of increasing bond length: 3 6 1 6 2 (c) In order of increasing bond enthalpy: 2 6 1 6 3 8.9 (a) Valence electrons are those that take part in chemical bonding. This usually means the electrons beyond the core noble-gas configuration of the atom, although it is sometimes only the outer-shell electrons. (b) A nitrogen atom has 5 valence electrons. (c) The atom (Si) has 4 valence electrons. 8.11 Si, 1s22s22p63s23p2. The n = 3 electrons are valence electrons; the others are nonvalence electrons. Valence electrons participate in chemical bonding; the others do not. 8.13 (a) Al 8.15 Mg O
(b) Br
(c) Ar Mg2
(d) Sr O
2
Answers to Selected Exercises 8.33
Cl
Cl Cl
Cl
Cl
Si
Cl
Si
Cl
the bond. Thus, the C ¬ O bond lengths vary in the order CO 6 CO2 6 CO32-. 8.57 (a) Two equally valid Lewis structures can be drawn for benzene.
8.47 (a) H
Si
(b) C
H
O
H
H
H
H
H
H
H
H
H H The concept of resonance dictates that the true description of bonding is some hybrid or blend of these two Lewis structures. The most obvious blend of these two resonance structures is a molecule with six equivalent C ¬ C bonds with equal lengths. (b) This model predicts a uniform C ¬ C bond length that is shorter than a single bond but longer than a double bond. 8.59 (a) The octet rule states that atoms will gain, lose, or share electrons until they are surrounded by eight valence electrons. (b) The octet rule applies to atoms in a covalent compound and the individual ions in an ionic compound. In the covalent compound CCl4, the atoms share electrons in order to surround themselves with an octet. In the ionic compound MgCl 2, Mg loses 2 e - to become Mg 2 + with the electron configuration of Ne. Each Cl atom gains one electron to form Cl- with the electron configuration of Ar. 8.61 No chlorine oxide will obey the octet rule. Chlorine has seven valence electrons, and oxygen has six. For neutral chlorine oxides, regardless of the number of oxygen atoms in the molecule, the total number of valence electrons will be an (odd + even) sum, which is always an odd number. 8.63
H
2
(a)
O
(c) F
S
F
(d) O
S
O
O
Cl
O
(f) H
N
O
S
(b) H
O
H
(c)
8.49 (a) Formal charge is the charge on each atom in a molecule, assuming all atoms have the same electronegativity. (b) Formal charges are not actual charges. They are a bookkeeping system that assumes perfect covalency, one extreme for the possible electron distribution in a molecule. (c) Oxidation numbers are a bookkeeping system that assumes the more electronegative element holds all electrons in a bond. The true electron distribution is some composite of the two extremes. 8.51 Formal charges are shown on the Lewis structures; oxidation numbers are listed below each structure.
N
N
N
O
C
S
0
0
0
(b)
O, 2; C, 4; S, 2
0
Cl
S
(d)
(e)
Cl Cl
C
1
S, 4; Cl, 1; O, 2 1
1
O
(c)
1
O
Br
O
1
2
Br, 5; O, 2
(d)
0
H
O
Cl
0
1
O
1
Cl, 3; H, 1; O, 2
8.53 (a) O N O O N O (b) O3 is isoelectronic with NO2-; both have 18 valence electrons. (c) Since each N ¬ O bond has partial double-bond character, the N ¬ O bond length in NO2- should be shorter than an N ¬ O single bond. 8.55 The more electron pairs shared by two atoms, the shorter
F
H
F
N
N
F
Sb
H
F
F
10 electrons around Sb 0
0
N
8.65 (a) Cl Cl
N
N
1
O
6 electrons around Al
N
H
H
Other resonance structures that minimize formal charges but violate the octet rule can be drawn. The octet rule versus formal charge debate is ongoing.
H
(a)
Al
O
H
O
H
O
(e)
H
H
Cl 8.35 (a) O O (b) A double bond is required because there are not enough electrons to satisfy the octet rule with single bonds and unshared pairs. (c) The greater the number of shared electron pairs between two atoms, the shorter the distance between the atoms. An O “ O double bond is shorter than an O ¬ O single bond. 8.37 (a) Electronegativity is the ability of an atom in a molecule to attract electrons to itself. (b) The range of electronegativities on the Pauling scale is 0.7–4.0. (c) Fluorine is the most electronegative element. (d) Cesium is the least electronegative element that is not radioactive. 8.39 (a) Mg (b) S (c) C (d) As 8.41 The bonds in (a), (c), and (d) are polar. The more electronegative element in each polar bond is (a) F (c) O (d) I. 8.43 (a) The calculated charge on H and Br is 0.12e. (b) From Sample Exercise 8.5, the calculated charge on H and Cl in HCl is 0.178e. HBr has a smaller dipole moment and longer bond length than HCl; these properties both contribute to the smaller charge separation in HBr. 8.45 (a) SiCl 4, molecular, silicon tetrachloride; LaF3, ionic, lanthanum(III) fluoride (b) FeCl 2, ionic, iron(II) chloride; ReCl 6, molecular (metal in high oxidation state), rhenium hexachloride. (c) PbCl 4, molecular (by contrast to the distinctly ionic RbCl), lead tetrachloride; RbCl, ionic, rubidium chloride H
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Be
Cl
0
0
This structure violates the octet rule. (b) Cl Be Cl Cl Be Cl 1
2
1
2
0
2
Cl
2
Be 2
Cl 0
(c) Formal charges are minimized on the structure that violates the octet rule; this form is probably dominant. 8.67 Three resonance structures for HSO3- are shown here. Because the ion has a 1- charge, the sum of the formal charges of the atoms is -1. 0 1 1 1 0 0 0 1 H O S O H O S O O
O
1
1
H
0
1
0
O
S
O
O 0
1
A-10
Answers to Selected Exercises
The structure with no double bonds obeys the octet rule for all atoms, but does not lead to minimized formal charges. The structures with one and two double bonds both minimize formal charge but do not obey the octet rule. Of these two, the structure with one double bond is preferred because the formal charge is localized on the more electronegative oxygen atom. 8.69 (a) ¢H = -304 kJ (b) ¢H = - 82 kJ (c) ¢H = -467 kJ 8.71 (a) ¢H = -321 kJ (b) ¢H = -103 kJ (c) ¢H = -203 kJ 8.73 (a) -97 kJ; exothermic (b) The ¢H calculated from bond enthalpies (-97 kJ) is slightly more exothermic (more negative) than that obtained using ¢H f° values (-92.38 kJ). 8.75 The average Ti ¬ Cl bond enthalpy is 430 kJ>mol. 8.77 (a) Six (nonradioactive) elements. Yes, they are in the same family, assuming H is placed with the alkali metals. The Lewis symbol represents the number of valence electrons of an element, and all elements in the same family have the same number of valence electrons. By definition of a family, all elements with the same Lewis symbol must be in the same family. 8.81 The charge on M is likely to be 3 + . The range of lattice energies for ionic compounds with the general formula MX and a charge of 2+ on the metal is 3-4 * 103 kJ>mol. The lattice energy of 6 * 103 kJ>mol indicates that the charge on M must be greater than 2+ . 8.85 (a) B ¬ O. The most polar bond will be formed by the two elements with the greatest difference in electronegativity. (b) Te ¬ I. These elements have the two largest covalent radii among this group. (c) TeI2. The octet rule is be satisfied for all three atoms. (d) P2O3 . Each P atom needs to share 3 e - and each O atom 2 e - to achieve an octet. And B2O3. Although this is not a purely ionic compound, it can be understood in terms of gaining and losing electrons to achieve a noble-gas configuration. If each B atom were to lose 3 e - and each O atom were to gain 2 e - , charge balance and the octet rule would be satisfied. 8.90 (a) +1 (b) -1 (c) +1 (assuming the odd electron is on N) (d) 0 (e) +3 8.95 An experimentally determined molecular structure will reveal bond lengths and angles of the B ¬ A “ B molecule. If resonance structures are important, the two B ¬ A bond lengths will be identical. If the molecule features one single and one double bond, the lengths will be significantly different. 8.98 (a) ¢H = 7.85 kJ>g nitroglycerine (b) 4 C7H5N3O6(s) ¡ 6 N2(g) + 7 CO2(g) + 10 H2O(g) + 21 C(s) 8.101 (a) Ti2 + , 3Ar43d2; Ca, 3Ar44s2. Yes. The 2 valence electrons in Ti2 + and Ca are in different principal quantum levels and different subshells. (b) In Ca the 4s is lower in energy than the 3d, while in Ti2 + the 3d is lower in energy than the 4s. (c) No. There is only one 4s orbital, so the 2 valence electrons in Ca are paired; there are 5 degenerate 3d orbitals, so the 2 valence electrons in Ti2 + are unpaired. 8.107 (a) Azide ion is N3- . (b) Resonance structures with formal charges are shown.
N
N
1
1
N 1
N 0
N 1
N 2
N
N
N
2
1
0
(c) The structure with two double bonds minimizes formal charges and is probably the main contributor. (d) The N ¬ N distances will be equal and have the approximate length of a N ¬ N double bond, 1.24 Å. 8.112 (a) D(Br ¬ Br)(l) = 223.6 kJ; D(Br ¬ Br)(g) = 193 kJ (b) D(C —Cl)(l) = 336.1 kJ; D(C ¬ Cl)(g) = 328 kJ (c) D(O ¬ O)(l) = 192.7 kJ; D(O ¬ O)(g) = 146 kJ (d) Breaking bonds in the liquid requires more energy than breaking bonds in the gas phase. Bond dissociation in the liquid phase can be thought of in two steps, vaporization of the liquid followed by bond dissociation in the gas phase. The greater bond dissociation enthalpy in the liquid phase is due to the contribution from the enthalpy of vaporization.
CHAPTER 9 9.1 Removing an atom from the equatorial plane of the trigonal bipyramid in Figure 9.3 creates a seesaw shape. 9.3 (a) 2 electron-domain geometries, linear and trigonal bipyramidal (b) 1 electron-domain
geometry, trigonal bipyramidal (c) 1 electron-domain geometry, octahedral (c) 1 electron-domain geometry, octahedral (d) 1 electron domain geometry, octahedral (e) 1 electron domain geometry, octahedral (f) 1 electron-domain geometry, trigonal bipyramidal (This triangular pyramid is an unusual molecular geometry not listed in Table 9.3. It could occur if the equatorial substituents on the trigonal bipyramid were extremely bulky, causing the nonbonding electron pair to occupy an axial position.) 9.5 (a) Zero energy corresponds to two separate, noninteracting Cl atoms. This infinite Cl ¬ Cl distance is beyond the right extreme of the horizontal axis on the diagram. (b) According to the valence bond model, valence orbitals on the approaching atoms overlap, allowing two electrons to mutually occupy space between the two nuclei and be stabilized by two nuclei rather than one. (c) The Cl ¬ Cl distance at the energy minimum on the plot is the Cl ¬ Cl bond length. (d) At interatomic separations shorter than the bond distance, the two nuclei begin to repel each other, increasing the overall energy of the system. (e) The y-coordinate of the minimum point on the plot is a good estimate of the Cl ¬ Cl bond energy or bond strength. 9.6 SiCl 4, 109°; PF3, 107°; SF2, 105°. Each molecule has tetrahedral electron domain geometry, but the number of nonbonding electron pairs increases from 0 to 2, respectively. Because nonbonding electron pairs occupy more space than bonding pairs, we expect the bond angles to decrease in the series. 9.9 (a) i, Two s atomic orbitals; ii, two p atomic orbitals overlapping end to end; iii, two p atomic orbitals overlapping side to side (b) i, s-type MO; ii, s-type MO; iii, p-type MO (c) i, antibonding; ii, bonding; iii, antibonding (d) i, the nodal plane is between the atom centers, perpendicular to the interatomic axis and equidistant from each atom. ii, there are two nodal planes; both are perpendicular to the interatomic axis. One is left of the left atom and the second is right of the right atom. iii, there are two nodal planes; one is between the atom centers, perpendicular to the interatomic axis and equidistant from each atom. The second contains the interatomic axis and is perpendicular to the first. 9.11 (a) Yes. The stated shape defines the bond angle and the bond length tells the size. (b) No. Atom A could have 2, 3, or 4 nonbonding electron pairs. 9.13 A molecule with tetrahedral molecular geometry has an atom at each vertex of the tetrahedron. A trigonal-pyramidal molecule has one vertex of the tetrahedron occupied by a nonbonding electron pair rather than an atom. 9.15 (a) An electron domain is a region in a molecule where electrons are most likely to be found. (b) Like the balloons in Figure 9.5, each electron domain occupies a finite volume of space, so they also adopt an arrangement where repulsions are minimized. 9.17 (a) The number of electron domains in a molecule or ion is the number of bonds (double and triple bonds count as one domain) plus the number of nonbonding electron pairs. (b) A bonding electron domain is a region between two bonded atoms that contains one or more pairs of bonding electrons. A nonbonding electron domain is localized on a single atom and contains one pair of nonbonding electrons. 9.19 (a) No effect on molecular shape (b) 1 nonbonding pair on P influences molecular shape (c) no effect (d) no effect (e) 1 nonbonding pair on S influences molecular shape 9.21 (a) 2 (b) 1 (c) none (d) 3 9.23 The electrondomain geometry indicated by VSEPR describes the arrangement of all bonding and nonbonding electron domains. The molecular geometry describes just the atomic positions. In H2O there are 4 electron domains around oxygen, so the electron-domain geometry is tetrahedral. Because there are 2 bonding and 2 nonbonding domains, the molecular geometry is bent. We make this distinction because all electron domains must be considered when describing the atomic arrangement and bond angles in a molecule but the molecular geometry or shape is a description of just the atom positions. 9.25 (a) Tetrahedral, tetrahedral (b) trigonal bipyramidal, T-shaped (c) octahedral, square pyramidal (d) octahedral, square planar 9.27 (a) Linear, linear (b) tetrahedral, trigonal pyramidal (c) trigonal bipyramidal, seesaw (d) octahedral, octahedral (e) tetrahedral, tetrahedral (f) linear, linear 9.29 (a) i, trigonal planar; ii, tetrahedral; iii, trigonal bipyramidal (b) i, 0; ii, 1; iii, 2 (c) N and P (d) Cl (or Br or I). This T-shaped molecular geometry arises from a trigonal-bipyramidal electron-domain geometry with 2 nonbonding domains. Assuming each F atom has 3 nonbonding domains and forms only single bonds with A, A must have 7 valence electrons and be in or below the third row of the periodic table
Answers to Selected Exercises to produce these electron-domain and molecular geometries. 9.31 (a) 1-109°, 2-109° (b) 3 -109°, 4 -109° (c) 5-180° (d) 6-120°, 7 -109°, 8 -109° 9.33 The two molecules with trigonal-bipyramidal electron-domain geometry, PF5 and SF4, have more than one F ¬ A ¬ F bond angle. 9.35 (a) Although both ions have 4 bonding electron domains, the 6 total domains around Br require octahedral domain geometry and square-planar molecular geometry, while the 4 total domains about B lead to tetrahedral domain and molecular geometry. (b) The less electronegative the central atom, the larger the nonbonding electron domain, and the greater the effect of repulsive forces on adjacent bonding domains. The less electronegative the central atom, the greater the deviation from ideal tetrahedral angles. The angles will vary as H2O 7 H2S 7 H2Se. 9.37 A bond dipole is the asymmetric charge distribution between two bonded atoms with unequal electronegativities. A molecular dipole moment is the three-dimensional sum of all the bond dipoles in a molecule. 9.39 (a) Yes. The net dipole moment vector points along the Cl ¬ S ¬ Cl angle bisector. (b) No, BeCl 2 does not have a dipole moment. 9.41 (a) In Exercise 9.29, molecules (ii) and (iii) will have nonzero dipole moments. Molecule (i) has no nonbonding electron pairs on A, and the 3 A ¬ F bond dipoles are oriented so that they cancel. Molecules (ii) and (iii) have nonbonding electron pairs on A and their bond dipoles do not cancel. (b) In Exercise 9.30, molecules (i) and (ii) have a zero dipole moment. 9.43 (a) IF (d) PCl 3 and (f) IF5 are polar. 9.45 (a) Lewis structures H Cl H H H Cl C
C
C
C
C
C
Cl
Cl
Cl H
H
Cl
Molecular geometries H
H C
H
C
Cl
Cl C
Cl
Cl
Polar
H
C
Cl C
H
C
H
Nonpolar
Cl Polar
(b) The middle isomer has a zero net dipole moment. (c) C2H3Cl has only one isomer, and it has a dipole moment. 9.47 (a) Orbital overlap occurs when valence atomic orbitals on two adjacent atoms share the same region of space. (b) A chemical bond is a concentration of electron density between the nuclei of two atoms. This concentration can take place because orbitals on the two atoms overlap. 9.49 (a) H ¬ Mg ¬ H, linear electron domain and molecular geometry (b) The linear electron-domain geometry in MgH2 requires sp hybridization. (c) H
hybridized to form three equivalent hybrid orbitals in a trigonal-planar arrangement. (b) sp2 (d) A single 2p orbital is unhybridized. It lies perpendicular to the trigonal plane of the sp2 hybrid orbitals. 9.55 (a) sp2 (b) sp3 (c) sp (d) sp3 9.57 No hybrid orbitals discussed in this chapter form angles of 90° with each other; p atomic orbitals are perpendicular to each other. 109.5°, sp3; 120°, sp2 9.59 (a) (b) (c) A s bond is generally stronger than a p bond because there is more extensive orbital overlap. (d) No. Overlap of two s orbitals results in electron density along the internuclear axis, while a p bond has none. 9.61 (a) H H H H H
C
C
H
H
C
H H
C
H
1s
CO2 NH3 CH4 BH3 SF4 SF6 H2CO PF5 XeF2
sp sp3 sp3 sp2 Not applicable Not applicable sp2 Not applicable Not applicable
Linear Tetrahedral Tetrahedral Trigonal planar Trigonal bipyramidal Octahedral Trigonal planar Trigonal bipyramidal Trigonal bipyramidal
No Yes No No Yes No Yes No No
9.53 (a) B, 3He42s22p1. One 2s electron is “promoted” to an empty 2p orbital. The 2s and two 2p orbitals that each contain one electron are
C
H
H
1s s1s
9.51 Hybridization Dipole Moment? of Central Atom Yes or No
C
(b) sp3, sp2, sp (c) nonplanar, planar, planar (d) 7 s, 0 p; 5 s, 1 p; 3 s, 2 p (e) The Si analogs would have the same hybridization as the C compounds given in part (b). That Si is in the row below C means it has a larger bonding atomic radius and atomic orbitals than C. The close approach of Si atoms required to form strong, stable p bonds in Si2H4 and Si2H2 is not possible and these Si analogs do not readily form. 9.63 (a) 18 valence electrons (b) 16 valence electrons form s bonds. (c) 2 valence electrons form p bonds. (d) No valence electrons are nonbonding. (e) The left and central C atoms are sp2 hybridized; the right C atom is sp3 hybridized. 9.65 (a) ' 109° about the leftmost C, sp3; ' 120° about the right-hand C, sp2 (b) The doubly bonded O can be viewed as sp2, and the other as sp3; the nitrogen is sp3 with approximately 109° bond angles. (c) nine s bonds, one p bond 9.67 (a) In a localized p bond, the electron density is concentrated between the two atoms forming the bond. In a delocalized p bond, the electron density is spread over all the atoms that contribute p orbitals to the network. (b) The existence of more than one resonance form is a good indication that a molecule will have delocalized p bonding. (c) delocalized 9.69 (a) Linear (b) The two central C atoms each have trigonal planar geometry with ' 120° bond angles about them. The C and O atoms lie in a plane with the H atoms free to rotate in and out of this plane. (c) The molecule is planar with ' 120° bond angles about the two N atoms. 9.71 (a) Hybrid orbitals are mixtures of atomic orbitals from a single atom and remain localized on that atom. Molecular orbitals are combinations of atomic orbitals from two or more atoms and are delocalized over at least two atoms. (b) Each MO can hold a maximum of two electrons. (c) Antibonding molecular orbitals can have electrons in them. 9.73 (a) s*1s s*1s
Mg
Electron-Domain Molecule Geometry
A-11
s1s
H2 (b) There is one electron in H2 + . (c) s11s (d) BO = 12 (e) Fall apart. If the single electron in H2 + is excited to the s*1s orbital, its energy is higher than the energy of an H 1s atomic orbital and H2 + will decompose into a hydrogen atom and a hydrogen ion. 9.75 z z y x
y x
(a) 1 s bond (b) 2 p bonds (c) 1 s* and 2 p* 9.77 (a) When comparing the same two bonded atoms, bond order and bond energy are
A-12
Answers to Selected Exercises
directly related, while bond order and bond length are inversely related. When comparing different bonded nuclei, there are no simple relationships. (b) Be2 is not expected to exist; it has a bond order of zero and is not energetically favored over isolated Be atoms. Be 2 + has a bond order of 0.5 and is slightly lower in energy than isolated Be atoms. It will probably exist under special experimental conditions. 9.79 (a, b) Substances with no unpaired electrons are weakly repelled by a magnetic field. This property is called diamagnetism. (c) O22- , Be 2 2 + 9.81 (a) B2 + , s2s2s*2s2p2p1, increase (b) Li2 + , s1s2s*1s2s2s1, increase (c) N2 + , s2s2s*2s2p2p4s2p1, increase (d) Ne 2 2 + , decrease 9.83 CN, s2s2s*2s2s2p2p2p4p*2p4, s2s2s*2s2s2p2p2p3, bond order = 2.5; CN + , s2s2s*2s2s2p2p2p2, bond order = 2.0; CN- , s2s2s*2s2s2p2p2p4, bond order = 3.0. (a) CN- (b) CN, CN + 9.85 (a) 3s, 3px, 3py, 3pz (b) p3p (c) 2 (d) If the MO diagram for P2 is similar to that of N2, P2 will have no unpaired electrons and be diamagnetic. 9.89 SiF4 is tetrahedral, SF4 is seesaw, XeF4 is square planar. The shapes are different because the number of nonbonding electron domains is different in each molecule, even though all have four bonding electron domains. Bond angles and thus molecular shape are determined by the total number of electron domains. 9.92 (a) 2 s bonds, 2 p bonds (b) 3 s bonds, 4 p bonds (c) 3 s bonds, 1 p bond (d) 4 s bonds, 1 p bond 9.94 BF3 is trigonal planar, the B ¬ F bond dipoles cancel and the molecule is nonpolar. PF3 has a tetrahedral electron-domain geometry with one position occupied by a nonbonding electron pair. The nonbonding electron pair ensures an asymmetrical electron distribution and the molecule is polar. 9.99 H H C C C H H (a) The molecule is nonplanar. (b) Allene has no dipole moment. (c) The bonding in allene would not be described as delocalized. The p electron clouds of the two adjacent C “ C are mutually perpendicular, so there is no overlap and no delocalization of p electrons. 9.101 (a) All O atoms have sp2 hybridization. (b) The two s bonds are formed by overlap of sp2 hybrid orbitals, the p bond is formed by overlap of atomic p orbitals, one nonbonded pair is in a p atomic orbital and the other five nonbonded pairs are in sp2 hybrid orbitals. (c) unhybridized p atomic orbitals (d) four, two from the p bond and two from the nonbonded pair in the p atomic orbital 9.104 s2s2s*2s2p2p4s2p1p*2p1 (a) Paramagnetic (b) The bond order of N2 in the ground state is 3; in the first excited state it has a bond order of 2. Owing to the reduction in bond order, N2 in the first excited state has a weaker N ¬ N bond. 9.110 (a) 2 SF4(g) + O2(g) ¡ 2 OSF4(g) (b) O F
S F
F F
(c) ¢H = -551 kJ, exothermic (d) The electron-domain geometry is trigonal bipyramidal. The O atom can be either equatorial or axial. (e) Since F is more electronegative than O, the structure that minimizes 90° F ¬ S ¬ F angles, the one with O axial, is preferred.
CHAPTER 10 10.1 It would be much easier to drink from a straw on Mars. When a straw is placed in a glass of liquid, the atmosphere exerts equal pressure inside and outside the straw. When we drink through a straw, we withdraw air, thereby reducing the pressure on the liquid inside. If only 0.007 atm is exerted on the liquid in the glass, a very small reduction in pressure inside the straw will cause the liquid to rise. 10.4 (a) As the reaction proceeds at constant temperature and pressure, the number of particles decreases and the container volume decreases. (b) As the reaction proceeds at constant volume and temperature, the number of particles decreases and pressure decreases. 10.7 (a) Pred 6 Pyellow 6 Pblue
(b) Pred = 0.28 atm; Pyellow = 0.42 atm; Pblue = 0.70 atm 10.10 (a) P(ii) 6 P(i) = P(iii) (b) PHe(iii) 6 PHe(ii) 6 PHe(i) (c) d(ii) 6 d(i) 6 d(iii) (d) The average kinetic energies of the particles in the three containers are equal. 10.13 (a) A gas is much less dense than a liquid. (b) A gas is much more compressible than a liquid. (c) All mixtures of gases are homogenous. Similar liquid molecules form homogeneous mixtures, while very dissimilar molecules form heterogeneous mixtures. (d) Both gases and liquids conform to the shape of their container. A gas also adopts the volume of its container, while a liquid maintains its own volume. 10.15 (a) 1.8 * 103 kPa (b) 18 atm (c) 2.6 * 102 lb>in.2 10.17 (a) 10.3 m (b) 2.1 atm 10.19 (a) The tube can have any cross-sectional area. (b) At equilibrium the force of gravity per unit area acting on the mercury column is not equal to the force of gravity per unit area acting on the atmosphere. (c) The column of mercury is held up by the pressure of the atmosphere applied to the exterior pool of mercury. (d) If you took the mercury barometer with you on a trip from the beach to high mountains, the height of the mercury column would decrease with elevation. 10.21 (a) 0.349 atm (b) 265 mm Hg (c) 3.53 * 104 Pa (d) 0.353 bar (e) 5.13 psi 10.23 (a) P = 773.4 torr (b) P = 1.018 atm (c) The pressure in Chicago is greater than standard atmospheric pressure, and so it makes sense to classify this weather system as a “high-pressure system.” 10.25 (i) 0.31 atm (ii) 1.88 atm (iii) 0.136 atm 10.27 (a) If V decreases by a factor of 4, P increases by a factor of 4. (b) If T decreases by a factor of 2, P decreases by a factor of 2. (c) If n decreases by a factor of 4, P decreases by a factor of 4. 10.29 (a) If equal volumes of gases at the same temperature and pressure contain equal numbers of molecules and molecules react in the ratios of small whole numbers, it follows that the volumes of reacting gases are in the ratios of small whole numbers. (b) Since the two gases are at the same temperature and pressure, the ratio of the numbers of atoms is the same as the ratio of volumes. There are 1.5 times as many Xe atoms as Ne atoms. (c) Yes. By definition, one mole of an ideal gas contains Avogadro’s number of particles. At a given temperature and pressure, equal numbers of particles occupy the same volume, so one mole of an ideal gas will always occupy the same volume at the given temperature and pressure. 10.31 (a) An ideal gas exhibits pressure, volume, and temperature relationships described by the equation PV = nRT. (b) Boyle’s law, V = constant>P; Charles’s law, V = constant * T; Avogadro’s law, V = constant * n. Collect all the equalities: V = (constant * T * n)>P. Call the constant R and multiply both sides of the equation by P, PV = nRT. (c) PV = nRT; P in atmospheres, V in liters, n in moles, T in kelvins. (d) R = 0.08315 L-bar/mol-K. 10.33 Flask A contains the gas with M = 30 g>mol, and flask B contains the gas with M = 60 g>mol. 10.35 P 2.00 atm 0.300 atm 650 torr 10.3 atm
V 1.00 L 0.250 L 11.2 L 585 mL
n
T
0.500 mol 3.05 * 10-3 mol 0.333 mol 0.250 mol
48.7 K 27 °C 350 K 295 K
10.37 8.2 * 102 kg He 10.39 (a) 5.15 * 1022 molecules (b) 6.5 kg air 10.41 (a) 91 atm (b) 2.3 * 102 L 10.43 (a) 29.8 g Cl2 (b) 9.42 L (c) 501 K (d) 2.28 atm 10.45 (a) n = 2 * 10-4 mol O2 (b) The roach needs 8 * 10-3 mol O2 in 48 h, approximately 100% of the O2 in the jar. 10.47 (a) 1.32 * 107 L (b) 5.1 * 108 mol Hg 10.49 For gas samples at the same conditions, molar mass determines density. Of the three gases listed, (c) Cl2 has the largest molar mass. 10.51 (c) Because the helium atoms are of lower mass than the average air molecule, the helium gas is less dense than air. The balloon thus weighs less than the air displaced by its volume. 10.53 (a) d = 1.77 g>L (b) M = 80.1 g>mol 10.55 M = 89.4 g>mol 10.57 4.1 * 10-9 g Mg 10.59 (a) 21.4 L CO2 (b) 40.7 L O2 10.61 0.402 g Zn 10.63 (a) When the stopcock is opened, the volume occupied by N2(g) increases from 2.0 L to 5.0 L. PN2 = 0.40 atm (b) When the gases mix, the volume of O2(g) increases from 3.0 L to 5.0 L. PO2 = 1.2 atm (c) Pt = 1.6 atm
Answers to Selected Exercises 10.65 (a) PHe = 1.87 atm, PNe = 0.807 atm, PAr = 0.269 atm, (b) Pt = 2.95 atm 10.67 xCO2 = 0.00039 10.69 PCO2 = 0.305 atm, Pt = 1.232 atm 10.71 PN2 = 1.3 atm, PO2 = 0.54 atm, PCO2 = 0.27 atm 10.73 2.5 mole % O2 10.75 Pt = 2.47 atm 10.77 (a) Increase in temperature at constant volume or decrease in volume or increase in pressure (b) decrease in temperature (c) increase in volume, decrease in pressure (d) increase in temperature 10.79 The fact that gases are readily compressible supports the assumption that most of the volume of a gas sample is empty space. 10.81 Average speed is the sum of the speeds of all particles divided by the total number of particles. The root mean square speed is the speed of a molecule with the same kinetic energy as the average kinetic energy of the sample. The root mean square speed is larger for a given gas sample at a fixed temperature, but the difference between the two is small. 10.83 (a) Average kinetic energy of the molecules increases. (b) Root mean square speed of the molecules increases. (c) Strength of an average impact with the container walls increases. (d) Total collisions of molecules with walls per second increases. 10.85 (a) In order of increasing speed and decreasing molar mass: HBr 6 NF3 6 SO2 6 CO 6 Ne (b) uNF3 = 324 m>s (c) The most probable speed of an ozone molecule in the stratosphere is 306 m/s. 10.87 Effusion is the escape of gas molecules through a tiny hole. Diffusion is the distribution of a gas throughout space or throughout another substance. 10.89 The order of increasing rate of effusion is 2H37Cl 6 1H37Cl 6 2H35Cl 6 1H35Cl. 10.91 As4S6 10.93 (a) Non-ideal-gas behavior is observed at very high pressures and low temperatures. (b) The real volumes of gas molecules and attractive intermolecular forces between molecules cause gases to behave nonideally. (c) According to the ideal-gas law, the ratio PV>RT should be constant for a given gas sample at all combinations of pressure, volume, and temperature. If this ratio changes with increasing pressure, the gas sample is not behaving ideally. 10.95 Ar (a = 1.34, b = 0.0322) will behave more like an ideal gas than CO2 (a = 3.59, b = 0.427) at high pressures. 10.97 (a) P = 4.89 atm (b) P = 4.69 atm (c) Qualitatively, molecular attractions are more important as the amount of free space decreases and the number of molecular collisions increases. Molecular volume is a larger part of the total volume as the container volume decreases. 10.99 From the value of b for Xe, the nonbonding radius is 2.72 Å. From Figure 7.6, the bonding atomic radius of Xe is 1.30 Å. We expect the bonding radius of an atom to be smaller than its nonbonding radius, but this difference is quite large. 10.101 V = 3.1 mm3 10.105 (a) 13.4 mol C3H8(g) (b) 1.47 * 103 mol C3H8(l) (c) The ratio of moles liquid to moles gas is 110. Many more molecules and moles of liquid fit in a container of fixed volume because there is much less space between molecules in the liquid phase. 10.108 Pt = 5.3 * 102 torr 10.111 42.2 g O2 10.115 T2 = 687 °C 10.120 (a) P(ideal) = 177 atm (b) P(van der Waals) = 187.4 atm (c) Under the conditions of this problem (large number of moles of gas), the correction for the real volume of molecules dominates. 10.123 (a) 44.58% C, 6.596% H, 16.44% Cl, 32.38% N (b) C8H14N5Cl (c) Molar mass of the compound is required in order to determine molecular formula when the empirical formula is known. 10.128 (a) 5.02 * 108 L CH3OH(l) (b) CH4(g) + 2 O2(g) ¡ CO2(g) + 2 H2O(l), ¢H° = -890.4 kJ; ¢H for combustion of the methane is -1.10 * 1013 kJ. CH3OH(l) + 3>2 O2(g) ¡ CO2(g) + 2 H2O(l), ¢H ° = -726.6 kJ; ¢H for combustion of the methanol is -9.00 * 1012 kJ. (c) The enthalpy change upon combustion of 1.00 L of CH4(l) is -2.59 * 104 kJ and for 1.00 L of CH3OH(l), -1.79 * 104 kJ. Clearly CH4(l) has the higher enthalpy of combustion per unit volume.
CHAPTER 11 11.1 The diagram best describes a liquid. The particles are close together, mostly touching, but there is no regular arrangement or order. This rules out a gaseous sample, where the particles are far apart, and a crystalline solid, which has a regular repeating structure in all three directions. 11.4 (a) In its final state, methane is a gas at 185 °C. 11.5 (a) 385 mm Hg (b) 22 °C (c) 47 °C 11.6 The stronger the inter-
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molecular forces, the higher the boiling point of a liquid. Propanol, CH3CH2CH2OH, has hydrogen bonding and the higher boiling point. 11.7 (a) Normal boiling point, 360 K; normal freezing point, 260 K (b) (i) gas (ii) solid (iii) liquid (c) The triple point is approximately 185 K at 0.45 atm. 11.9 (a) Solid 6 liquid 6 gas (b) gas 6 liquid 6 solid (c) Matter in the gaseous state is most easily compressed because particles are far apart and there is much empty space. 11.11 Ar 6 CCl4 6 Si 11.13 (a) The molar volumes of Cl2 and NH3 are nearly the same because they are both gases. (b) On cooling to 160 K, both compounds condense from the gas phase to the solid-state, so we expect a significant decrease in the molar volume. (c) The molar volumes are 0.0351 L>mol Cl 2 and 0.0203 L>mol NH3 (d) Solid-state molar volumes are not as similar as those in the gaseous state, because most of the empty space is gone and molecular characteristics determine properties. Cl2(s) is heavier, has a longer bond distance and weaker intermolecular forces, so it has a significantly larger molar volume than NH3(s). (e) There is little empty space between molecules in the liquid state, so we expect their molar volumes to be closer to those in the solid state than those in the gaseous state. 11.15 (a) London dispersion forces (b) dipole–dipole and London dispersion forces (c) dipole–dipole forces and in certain cases hydrogen bonding 11.17 (a) SO2, dipole–dipole and London dispersion forces (b) CH3COOH, London dispersion, dipole–dipole, and hydrogen bonding (c) H2Se, dipole–dipole and London dispersion forces (but not hydrogen bonding) 11.19 (a) Polarizability is the ease with which the charge distribution in a molecule can be distorted to produce a transient dipole. (b) Sb is most polarizable because its valence electrons are farthest from the nucleus and least tightly held. (c) in order of increasing polarizability: CH4 6 SiH4 6 SiCl4 6 GeCl4 6 GeBr4 (d) The magnitudes of London dispersion forces and thus the boiling points of molecules increase as polarizability increases. The order of increasing boiling points is the order of increasing polarizability given in (c). 11.21 (a) H2S (b) CO2 (c) GeH4 11.23 Both rodlike butane molecules and spherical 2-methylpropane molecules experience dispersion forces. The larger contact surface between butane molecules facilitates stronger forces and produces a higher boiling point. 11.25 (a) A molecule must contain H atoms, bound to either N, O, or F atoms, in order to participate in hydrogen bonding with like molecules. (b) CH3NH2 and CH3OH 11.27 (a) Replacing a hydroxyl hydrogen with a CH3 group eliminates hydrogen bonding in that part of the molecule. This reduces the strength of intermolecular forces and leads to a lower boiling point. (b) CH3OCH2CH2OCH3 is a larger, more polarizable molecule with stronger London dispersion forces and thus a higher boiling point. 11.29 Physical Property Normal boiling point, °C Normal melting point, °C
H2O
H2S
100.00 0.00
-60.7 -85.5
(a) Based on its much higher normal melting point and boiling point, H2O has much stronger intermolecular forces. H2O has hydrogen bonding, while H2S has dipole–dipole forces. (b) H2S is probably a typical compound with less empty space in the ordered solid than the liquid, so that the solid is denser than the liquid. For H2O, maximizing the number of hydrogen bonds to each molecule in the solid requires more empty space than in the liquid, and the solid is less dense. (c) Specific heat is the energy required to raise the temperature of one gram of the substance one degree Celsius. Hydrogen bonding in water is such a strong attractive interaction that the energy required to disrupt it and increase molecular motion is large. 11.31 SO42- has a greater negative charge than BF4-, so ion–ion electrostatic attractions are greater in sulfate salts and they are less likely to form liquids. 11.33 (a) As temperature increases, the number of molecules with sufficient kinetic energy to overcome intermolecular attractive forces increases, and viscosity and surface tension decrease. (b) The same attractive forces that cause surface molecules to be difficult to separate
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Answers to Selected Exercises
(high surface tension) cause molecules elsewhere in the sample to resist movement relative to each other (high viscosity). 11.35 (a) CHBr3 has a higher molar mass, is more polarizable, and has stronger dispersion forces, so the surface tension is greater. (b) As temperature increases, the viscosity of the oil decreases because the average kinetic energy of the molecules increases. (c) Adhesive forces between polar water and nonpolar car wax are weak, so the large surface tension of water draws the liquid into the shape with the smallest surface area, a sphere. (d) Adhesive forces between nonpolar oil and nonpolar car wax are similar to cohesive forces in oil, so the oil drops spread out on the waxed car hood. 11.37 (a) The three molecules have similar structures and experience the same types of intermolecular forces. As molar mass increases, the strength of dispersion forces increases and the boiling points, surface tension, and viscosities all increase. (b) Ethylene glycol has an ¬ OH group at both ends of the molecule. This greatly increases the possibilities for hydrogen bonding; the overall intermolecular attractive forces are greater and the viscosity of ethylene glycol is much greater. (c) Water has the highest surface tension but lowest viscosity because it is the smallest molecule in the series. There is no hydrocarbon chain to inhibit their strong attraction to molecules in the interior of the drop, resulting in high surface tension. The absence of an alkyl chain also means the molecules can move around each other easily, resulting in the low viscosity. 11.39 (a) Melting, endothermic (b) evaporation, endothermic (c) deposition, exothermic (d) condensation, exothermic 11.41 Melting does not require separation of molecules, so the energy requirement is smaller than for vaporization, where molecules must be separated. 11.43 2.3 * 103 g H2O 11.45 (a) 39.3 kJ (b) 60 kJ 11.47 (a) The critical pressure is the pressure required to cause liquefaction at the critical temperature. (b) As the force of attraction between molecules increases, the critical temperature of the compound increases. (c) All the gases in Table 11.5 can be liquefied at the temperature of liquid nitrogen, given sufficient pressure. 11.49 (a) No effect (b) no effect (c) Vapor pressure decreases with increasing intermolecular attractive forces because fewer molecules have sufficient kinetic energy to overcome attractive forces and escape to the vapor phase. (d) Vapor pressure increases with increasing temperature because average kinetic energies of molecules increase. (e) Vapor pressure decreases with increasing density because attractive intermolecular forces increase. 11.51 (a) CBr4 6 CHBr3 6 CH2Br2 6 CH2Cl2 6 CH3Cl 6 CH4. The trend is dominated by dispersion forces even though four of the molecules are polar. The order of increasing volatility is the order of increasing vapor pressure, decreasing molar mass, and decreasing strength of dispersion forces. (b) Boiling point increases as the strength of intermolecular forces increases; this is the order of decreasing volatility and the reverse of the order in part (a). CH4 6 CH3Cl 6 CH2Cl2 6 CH2Br2 6 CHBr3 6 CBr4 11.53 (a) The temperature of the water in the two pans is the same. (b) Vapor pressure does not depend on either volume or surface area of the liquid. At the same temperature, the vapor pressures of water in the two containers are the same. 11.55 (a) Approximately 48 °C (b) approximately 340 torr (c) approximately 17 °C (d) approximately 1000 torr 11.57 (a) The critical point is the temperature and pressure beyond which the gas and liquid phases are indistinguishable. (b) The line that separates the gas and liquid phases ends at the critical point because at conditions beyond the critical temperature and pressure, there is no distinction between gas and liquid. In experimental terms a gas cannot be liquefied at temperatures higher than the critical temperature, regardless of pressure. 11.59 (a) H2O(g) will condense to H2O(s) at approximately 4 torr; at a higher pressure, perhaps 5 atm or so, H2O(s) will melt to form H2O(l). (b) At 100 °C and 0.50 atm, water is in the vapor phase. As it cools, water vapor condenses to the liquid at approximately 82 °C, the temperature where the vapor pressure of liquid water is 0.50 atm. Further cooling results in freezing at approximately 0 °C. The freezing point of water increases with decreasing pressure, so at 0.50 atm the freezing temperature is very slightly above 0 °C. 11.61 (a) 24 K (b) Neon sublimes at pressures less than the triple point pressure, 0.43 atm. (c) No 11.63 (a) Methane on the surface of Titan is likely to exist in both solid and liquid forms. (b) As
pressure decreases upon moving away from the surface of Titan, CH4(l) (at -178 °C) will vaporize to CH4(g), and CH4(s) (at temperatures below -180 °C) will sublime to CH4(g). 11.65 In a nematic liquid crystalline phase, molecules are aligned along their long axes, but the molecular ends are not aligned. Molecules are free to translate in all dimensions, but they cannot tumble or rotate out of the molecular plane, or the order of the nematic phase is lost and the sample becomes an ordinary liquid. In an ordinary liquid, molecules are randomly oriented and free to move in any direction. 11.67 The presence of polar groups or nonbonded electron pairs leads to relatively strong dipole–dipole interactions between molecules. These are a significant part of the orienting forces necessary for liquid crystal formation. 11.69 Because order is maintained in at least one dimension, the molecules in a liquid-crystalline phase are not totally free to change orientation. This makes the liquid-crystalline phase more resistant to flow, more viscous, than the isotropic liquid. 11.71 Melting provides kinetic energy sufficient to disrupt molecular alignment in one dimension in the solid, producing a smectic phase with ordering in two dimensions. Additional heating of the smectic phase provides kinetic energy sufficient to disrupt alignment in another dimension, producing a nematic phase with onedimensional order. 11.73 (a) Decrease (b) increase (c) increase (d) increase (e) increase (f) increase (g) increase 11.77 When a halogen is substituted for H in benzene, molar mass, polarizability and strength of dispersion forces increase; the order of increasing molar mass is the order of increasing boiling points for the first three compounds. C6H5OH experiences hydrogen bonding, the strongest force between neutral molecules, so it has the highest boiling point. 11.82 (a) Evaporation is an endothermic process. The heat required to vaporize sweat is absorbed from your body, helping to keep it cool. (b) The vacuum pump reduces the pressure of the atmosphere above the water until atmospheric pressure equals the vapor pressure of water and the water boils. Boiling is an endothermic process, and the temperature drops if the system is not able to absorb heat from the surroundings fast enough. As the temperature of the water decreases, the water freezes. 11.86 At low Antarctic temperatures, molecules in the liquid crystalline phase have less kinetic energy due to temperature, and the applied voltage may not be sufficient to overcome orienting forces among the ends of molecules. If some or all of the molecules do not rotate when the voltage is applied, the display will not function properly. 11.90 Br CH2
CH2 CH3 CH3
CH3
(i) M 44
CH2
CH CH3 CH3
CH2
CH3
(iii) M 123
(ii) M 72
O C CH3
CH2 CH3
(iv) M 58
CH3
Br CH2
(v) M 123
CH2 CH3
OH CH2
(vi) M 60
(a) Molar mass: Compounds (i) and (ii) have similar rodlike structures. The longer chain in (ii) leads to greater molar mass, stronger London dispersion forces, and higher heat of vaporization. (b) Molecular shape: Compounds (iii) and (v) have the same chemical formula and molar mass but different molecular shapes. The more rodlike shape of (v) leads to more contact between molecules, stronger dispersion forces, and higher heat of vaporization. (c) Molecular polarity: Compound (iv) has a smaller molar mass than (ii) but a larger heat of vaporization, which must be due to the presence of dipole–dipole forces. (d) Hydrogen bonding interactions: Molecules (v) and (vi) have similar structures. Even though (v) has larger molar mass and dispersion forces, hydrogen bonding causes (vi) to have the higher heat of vaporization. 11.93 P(benzene vapor) = 98.7 torr
Answers to Selected Exercises
CHAPTER 12 12.1 Two-dimensional structure
(i)
(ii)
(a) unit cell
(b) lattice (c) cell contents
square one black, one white
square one black
12.3 (a) 1 Re and 3 O atoms per unit cell (b) 3.92 Å (c) 6.46 g>cm3 12.5 We expect linear polymer (a), with ordered regions, to be denser and have a higher melting point than branched polymer (b). 12.7 In molecular solids, relatively weak intermolecular forces bind the molecules in the lattice, so relatively little energy is required to disrupt these forces. In covalent-network solids, covalent bonds join atoms into an extended network. Melting or deforming a covalent-network solid means breaking covalent bonds, which requires a large amount of energy. 12.9 (a) Hydrogen bonding, dipole-dipole forces, London dispersion forces (b) covalent chemical bonds (c) ionic bonds (d) metallic bonds 12.11 (a) Ionic (b) metallic (c) covalent-network (It could also be characterized as ionic with some covalent character to the bonds.) (d) molecular (e) molecular (f) molecular 12.13 Because of its relatively high melting point and properties as a conducting solution, the solid must be ionic. 12.15
Crystalline
Amorphous
12.17 Two-dimensional structure (a) unit cell
(b) g, a, b (c) lattice type
(i) A
B
l = 90°, a = b square
(ii) A
B
l = 120°, a = b hexagonal
12.19 (a) Orthorhombic (d) tetragonal 12.21 Triclinic, rhombohedral 12.23 There is a minimum of two (metal) atoms in a bodycentered cubic unit cell. 12.25 (a) Primitive hexagonal unit cell (b) NiAs 12.27 Moving left to right in the period, atomic mass and Zeff increase. The increase in Zeff leads to smaller bonding atomic radii and atomic volume. Mass increases, volume decreases, and density increases in the series. The variation in densities reflects shorter metal–metal bond distances and an increase in the extent of metal–metal bonding. The strength of metal–metal bonds in the series
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is probably the most important factor influencing the increase in density. 12.29 (a) Structure types A and C have equally dense packing and are more densely packed than structure type B. (b) Structure type B is least densely packed. 12.31 (a) The radius of an Ir atom is 1.355 Å. (b) The density of Ir is 22.67 g>cm3 12.33 (a) 4 Al atoms per unit cell (b) coordination number = 12 (c) a = 4.04 Å or 4.04 * 10-8 cm (d) density = 2.71 g>cm3 12.35 An alloy contains atoms of more than one element and has the properties of a metal. In a solution alloy the components are randomly dispersed. In a heterogeneous alloy the components are not evenly dispersed and can be distinguished at a macroscopic level. In an intermetallic compound the components have interacted to form a compound substance, as in Cu3As. 12.37 (a) Interstitial alloy (b) substitutional alloy (c) intermetallic compound 12.39 (a) True (b) false (c) false 12.41 (a) Nickel or palladium, substitutional alloy (b) copper, substitutional alloy (c) indium, intermetallic compound (d) silver, substitutional alloy 12.43 In the electron-sea model, valence electrons move about the metallic lattice, while metal atoms remain more or less fixed in position. Under the influence of an applied potential, the electrons are free to move throughout the structure, giving rise to thermal and electrical conductivity. 12.45
E
(a) Six AOs require six MOs (b) zero nodes in the lowest energy orbital (c) five nodes in highest energy orbital (d) two nodes in the HOMO (e) three nodes in the LUMO. 12.47 (a) Ag (b) Zn. Ductility decreases as the strength of metal–metal bonding increases, producing a stiffer lattice, less susceptible to distortion. 12.49 Moving from Y to Mo, the number of valence electrons, occupancy of the bonding band, and strength of metallic bonding increase. Stronger metallic bonding requires more energy to break bonds and mobilize atoms, resulting in higher melting points. 12.51 (a) SrTiO3 (b) Each Sr atom is coordinated to twelve O atoms in eight unit cells that contain the Sr atom. 12.53 (a) a = 4.70 Å (b) 2.69 g>cm3 12.55 (a) 7.711 g>cm3 (b) We expect Se2- to have a larger ionic radius than S2- , so HgSe will occupy a larger volume and the unit cell edge will be longer. (c) The density of HgSe is 8.241 g>cm3. The greater mass of Se accounts for the greater density of HgSe. 12.57 (a) Cs + and I- have the most similar radii and will adopt the CsCl-type structure. The radii of Na+ and I- are somewhat different; NaI will adopt the NaCl-type structure. The radii of Cu + and I- are very different; CuI has the ZnS-type structure. (b) CsI, 8; NaI, 6; CuI, 4 12.59 (a) 6 (b) 3 (c) 6 12.61 (a) False (b) true 12.63 (a) Ionic solids are much more likely to dissolve in water. (b) Covalent-network solids can become electrical conductors via chemical substitution. 12.65 (a) CdS (b) GaN (c) GaAs 12.67 Ge or Si (Ge is closer to Ga in bonding atomic radius.) 12.69 (a) A 1.1 eV photon corresponds to a wavelength of 1.1 * 10-6 m. (b) According to the figure, Si can absorb a portion of the visible light that comes from the sun. 12.71 l = 560 nm 12.73 The band gap is approximately 1.85 eV, which corresponds to a wavelength of 672 nm. 12.75 Monomers are small molecules with low molecular mass that are joined together to form polymers. Three monomers mentioned in this chapter are
A-16 CH2
Answers to Selected Exercises CH
CH2
CH
CH3
H C
CH3
C
CH2
Propylene (propene)
Styrene (phenyl ethene)
CH2
Isoprene (2-methyl-1,3-butadiene)
longer bond distance because the valence electrons are shared with twelve nearest neighbors rather than being localized in four bonds as in gray tin. 12.103 F F C
C
F
F
or n F
F 12.77
F
F
F
F C
C
C O CH3
C
O
H
H
CH2
O
Acetic acid
F
Ethanol
O C
CH3
O
CH2CH3
H2O
Ethyl acetate If a dicarboxylic acid and a dialcohol are combined, there is the potential for propagation of the polymer chain at both ends of both monomers. 12.79 (a) H H C
C
H (b)
CI CH2
CH2 CH2
H2N HO
CH2 CH2
CH2 C
NH2 CH2
CH2
O OH
O (c)
O OH
CH2 HO
12.81 HOOC
CH2
COOH
O C
HO
H2N
F
F
F
F
C F
F
F
Teflon ™ is formed by addition polymerization. 12.105 Diffraction, the phenomenon that enables us to measure interatomic distances in crystals, is most efficient when the wavelength of light is similar to or smaller than the size of the object doing the diffracting. Atom sizes are on the order of 1–10 Å, and the wavelengths of X-rays are also in this range. Visible light, 400–700 nm or 4000–7000 Å, is too long to be diffracted effectively by atoms (electrons) in crystals. 12.107 In a diffraction experiment, we expect a Ge crystal to diffract X-rays at a smaller θ angle than a Si crystal, assuming the X-rays have the same wavelength. 12.109 (a) The bonds in a semiconductor will be weakened by n-type doping. (b) The bonds in a semiconductor will also be weakened by p-type doping. 12.112 (a) 109° (b) 120° 12.113 (a) ¢H = -82 kJ>mol (b) ¢H = -14 kJ>mol (of either reactant) (c) ¢H = 0 kJ
CHAPTER 13
C
CH2 CH2
C
C
C
CH3
C OH
NH2
and 12.83 Flexibility of molecular chains causes flexibility of the bulk polymer. Flexibility is enhanced by molecular features that inhibit order, such as branching, and diminished by features that encourage order, such as cross-linking or delocalized p electron density. Crosslinking, the formation of chemical bonds between polymer chains, reduces flexibility of the molecular chains, increases the hardness of the material, and decreases the chemical reactivity of the polymer. 12.85 No. The function of the polymer determines whether high molecular mass and high degree of crystallinity are desirable properties. If the polymer will be used as a flexible wrapping or fiber, rigidity that is due to high molecular mass is an undesirable property. 12.87 If a solid has nanoscale dimensions of 1–10 nm, there may not be enough atoms contributing atomic orbitals to produce continuous energy bands of molecular orbitals. 12.89 (a) False. As particle size decreases, the band gap increases. (b) False. As particle size decreases, wavelength decreases. 12.91 2.47 * 105 Au atoms 12.94 The face-centered structure will have the greater density. 12.98 (a) CsCl, primitive cubic lattice (b) Au, face-centered cubic lattice (c) NaCl, facecentered cubic lattice (d) Po, primitive cubic lattice, rare for metals (e) ZnS, face-centered cubic lattice 12.99 White tin has a structure characteristic of a metal, while gray tin has the diamond structure characteristic of group 4A semiconductors. Metallic white tin has the
13.1 (a) 6 (b) 6 (c) 13.3 The greater the lattice energy of an ionic solid, the more endothermic the dissolving process and the less soluble the salt in water. 13.7 Vitamin B6 is largely water soluble because of its small size and capacity for extensive hydrogen-bonding interactions. Vitamin E is largely fat soluble. The long, rodlike hydrocarbon chain will lead to strong dispersion forces among vitamin E and mostly nonpolar fats. 13.9 (a) Yes, the molarity changes with a change in temperature. Molarity is defined as moles solute per unit volume of solution. A change of temperature changes solution volume and molarity. (b) No, molality does not change with change in temperature. Molality is defined as moles solute per kilogram of solvent. Temperature affects neither mass nor moles. 13.13 If the magnitude of ¢Hmix is small relative to the magnitude of ¢Hsolute, ¢Hsoln will be large and endothermic (energetically unfavorable) and not much solute will dissolve. 13.15 (a) Dispersion (b) hydrogen bonding (c) ion–dipole (d) dipole–dipole 13.17 Very soluble. In order for ¢Hsoln to be negative, ¢Hmix must have a greater magnitude than (¢Hsolute + ¢Hsolvent). The entropy of mixing always encourages solubility. In this case, the enthalpy of the system decreases and the entropy increases, so the ionic compound dissolves. 13.19 (a) ¢Hsolute (b) ¢Hmix 13.21 (a) Since the solute and solvent experience very similar London dispersion forces, the energy required to separate them individually and the energy released when they are mixed are approximately equal. ¢Hsolute + ¢Hsolvent L - ¢Hmix. Thus, ¢Hsoln is nearly zero. (b) Since no strong intermolecular forces prevent the molecules from mixing, they do so spontaneously because of the increase in randomness. 13.23 (a) Supersaturated (b) Add a seed crystal. A seed crystal provides a nucleus of prealigned molecules, so that ordering of the dissolved particles (crystallization) is more facile. 13.25 (a) Unsaturated (b) saturated (c) saturated (d) unsaturated 13.27 The liquids water and glycerol form homogenous mixtures (solutions) regardless of the relative amounts of the two components. The ¬ OH groups of glycerol facilitate strong hydrogen bonding similar to that in water; like dissolves like. 13.29 Toluene, C6H5CH3, is the best solvent for nonpolar solutes. Without polar groups or nonbonding electron pairs, it forms only dispersion interactions with itself and other molecules. 13.31 (a) Dispersion interactions among nonpolar CH3(CH2)16 ¬ chains dominate the properties of stearic acid, causing
Answers to Selected Exercises
3.5 * 10-2 m (b) Molality and molarity are numerically similar when kilograms solvent and liters solution are nearly equal. This is true when solutions are dilute and when the density of the solvent is nearly 1 g>mL, as in this exercise. (c) Water is a polar solvent; the solubility of solutes increases as their polarity increases. Nonpolar CF4 has the lowest solubility and the most polar fluorocarbon, CHClF2, has the greatest solubility in H2O. (d) The Henry’s law constant for CHClF2 is 3.5 * 10-2 mol>L-atm. This value is greater than the Henry’s law constant for N2(g) because N2(g) is nonpolar and of lower molecular mass than CHClF2. 13.109 (a) cation (g) anion (g) solvent Solvation energy of gaseous ions U lattice energy
Solution Hsoln
Ionic solid solvent (b) Lattice energy (U) is inversely related to the distance between ions, so salts with large cations like (CH3)4N + have smaller lattice energies than salts with simple cations like Na + . Also the ¬ CH3 groups in the large cation are capable of dispersion interactions with nonpolar groups of the solvent molecules, resulting in a more negative solvation energy of the gaseous ions. Overall, for salts with larger cations, lattice energy is smaller (less positive), the solvation energy of the gaseous ions is more negative, and ¢Hsoln is less endothermic. These salts are more soluble in polar nonaqueous solvents. 13.112 The freezing point of the LiOH(aq) solution is essentially zero, Tf = -0.00058 °C.
CHAPTER 14 14.1 Vessel 2 14.3 Equation (d) 14.9 (1) Total potential energy of the reactants (2) Ea, activation energy of the reaction (3) ¢E, net energy change for the reaction (4) total potential energy of the products 14.12 (a) NO2 + F2 ¡ NO2F + F; NO2 + F ¡ NO2F (b) 2 NO2 + F2 ¡ 2 NO2F (c) F (atomic fluorine) is the intermediate (d) rate = k3NO243F24) 14.16 Transition state (2)
Potential energy
it to be more soluble in nonpolar CCl4. (b) Dioxane can act as a hydrogen bond acceptor, so it will be more soluble than cyclohexane in water. 13.33 (a) CCl4 is more soluble because dispersion forces among nonpolar CCl4 molecules are similar to dispersion forces in hexane. (b) C6H6 is a nonpolar hydrocarbon and will be more soluble in the similarly nonpolar hexane. (c) The long, rodlike hydrocarbon chain of octanoic acid forms strong dispersion interactions and causes it to be more soluble in hexane. 13.35 (a) A sealed container is required to maintain a partial pressure of CO2(g) greater than 1 atm above the beverage. (b) Since the solubility of gases increases with decreasing temperature, more CO2(g) will remain dissolved in the beverage if it is kept cool. 13.37 SHe = 5.6 * 10-4 M, SN2 = 9.0 * 10-4 M 13.39 (a) 2.15% Na2SO4 by mass (b) 3.15 ppm Ag 13.41 (a) XCH3OH = 0.0427 (b) 7.35% CH3OH by mass (c) 2.48 m CH3OH 13.43 (a) 1.46 * 10-2 M Mg(NO3)2 (b) 1.12 M LiClO4 # 3 H2O (c) 0.350 M HNO3 13.45 (a) 4.70 m C6H6 (b) 0.235 m NaCl 13.47 (a) 43.01% H2SO4 by mass (b) XH2SO4 = 0.122 (c) 7.69 m H2SO4 (d) 5.827 M H2SO4 13.49 (a) XCH3OH = 0.227 (b) 7.16 m CH3OH (c) 4.58 M CH3OH 13.51 (a) 0.150 mol SrBr2 (b) 1.56 * 10-2 mol KCl (c) 4.44 * 10-2 mol C6H12O6 13.53 (a) Weigh out 1.3 g KBr, dissolve in water, dilute with stirring to 0.75 L. (b) Weigh out 2.62 g KBr, dissolve it in 122.38 g H2O to make exactly 125 g of 0.180 m solution. (c) Dissolve 244 g KBr in water, dilute with stirring to 1.85 L. (d) Weigh 10.1 g KBr, dissolve it in a small amount of water, and dilute to 0.568 L. 13.55 71% HNO3 by mass 13.57 (a) 3.82 m Zn (b) 26.8 M Zn 13.59 1.8 * 10-3 M CO2 13.61 Freezing point depression, ¢Tf = Kf(m); boiling-point elevation, ¢Tb = Kb(m); osmotic pressure, P = MRT; vapor pressure lowering, PA = XAPA° 13.63 (a) Sucrose has a greater molar mass than glucose, so the sucrose solution will contain fewer particles and have a higher vapor pressure. 13.65 (a) PH2O = 186.4 torr (b) 78.9 g C3H8O2 13.67 (a) XEth = 0.2812 (b) Psoln = 238 torr (c) XEth in vapor = 0.472 13.69 (a) Because NaCl is a strong electrolyte, one mole of NaCl produces twice as many dissolved particles as one mole of the molecular solute C6H12O6. Boiling-point elevation is directly related to total moles of dissolved particles, so 0.10 m NaCl has the higher boiling point. (b) 0.10 m NaCl: ¢Tb = 0.101 °C, Tb = 100.1 °C; 0.10 m C6H12O6: ¢Tb = 0.051 °C, Tb = 100.1 °C (c) Interactions between ions in solution result in nonideal behavior. 13.71 0.050 m LiBr 6 0.120 m glucose 6 0.050 m Zn(NO3)2 13.73 (a) Tf = -115.0 °C, Tb = 78.7 °C (b) Tf = -67.3 °C, Tb = 64.2 °C (c) Tf = -0.4 °C, Tb = 100.1 °C (c) Tf = -0.6 °C, Tb = 100.2 °C 13.75 167 g C2H6O2 13.77 P = 0.0168 atm = 12.7 torr 13.79 Experimental molar mass of adrenaline is 1.8 * 102 g. The structure shows a molecular formula of C9H13NO3, with a molar mass of 183 g. The two values agree to two significant figures, the precision of the experiment. 13.81 Molar mass of lysozyme = 1.39 * 104 g 13.83 (a) i = 2.8 (b) The more concentrated the solution, the greater the ion pairing and the smaller the measured value of i. 13.85 (a) In the gaseous state, particles are far apart and intermolecular attractive forces are small. When two gases combine, all terms in Equation 13.1 are essentially zero and the mixture is always homogeneous. (b) To determine whether Faraday’s dispersion is a true solution or a colloid, shine a beam of light on it. If light is scattered, the dispersion is a colloid. 13.87 (a) Hydrophobic (b) hydrophilic (c) hydrophobic (d) hydrophobic (but stabilized by adsorbed charges). 13.89 When electrolytes are added to a suspension of proteins, dissolved ions form ion pairs with the protein surface charges, effectively neutralizing them. The protein’s capacity for ion–dipole interactions with water is diminished and the colloid separates into a protein layer and a water layer. 13.91 The periphery of the BHT molecule is mostly hydrocarbon-like groups, such as ¬ CH3. The one ¬ OH group is rather buried inside and probably does little to enhance solubility in water. Thus, BHT is more likely to be soluble in the nonpolar hydrocarbon hexane, C6H14, than in polar water. 13.94 (a) kRn = 7.27 * 10-3 mol>L-atm (b) PRn = 1.1 * 10-4 atm; SRn = 8.1 * 10-7 M 13.98 (a) 2.69 m LiBr (b) XLiBr = 0.0994 (c) 81.1% LiBr by mass 13.100 XH2O = 0.808; 0.0273 mol ions; 0.0136 mol NaCl 13.103 (a) -0.6 °C (b) -0.4 °C 13.106 (a) CF4, 1.7 * 10-4 m; CClF3, 9 * 10-4 m; CCl2F2, 2.3 * 10-2 m; CHClF2,
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Transition state (1)
Ea(2) Transition state (3)
Ea(1) Reactants
Ea(3)
Intermediate (1)
Intermediate (2)
Products
Reaction progress 14.17 (a) Reaction rate is the change in the amount of products or reactants in a given amount of time. (b) Rates depend on concentration of reactants, surface area of reactants, temperature, and presence of catalyst. (c) The stoichiometry of the reaction (mole ratios of reactants and products) must be known to relate rate of disappearance of reactants to rate of appearance of products.
A-18
Answers to Selected Exercises
14.19 Time (min)
Mol A
(a) Mol B
[A] (mol>L)
¢[A] (mol>L)
(b) Rate (M>s)
0 10 20 30 40
0.065 0.051 0.042 0.036 0.031
0.000 0.014 0.023 0.029 0.034
0.65 0.51 0.42 0.36 0.31
-0.14 -0.09 -0.06 -0.05
2.3 1.5 1.0 0.8
* * * *
10-4 10-4 10-4 10-4
energetic. (c) The rate constant usually increases with an increase in reaction temperature. 14.55 f = 4.94 * 10-2. At 400 K approximately 1 out of 20 molecules has this kinetic energy. 14.57 (a) Ea 7 kJ
E 66 kJ
(c) ¢3B4avg>¢t = 1.3 * 10-4 M>s 14.21 (a) Time Concentration Time (s) Interval (s) (M) ¢M
Rate (M>s)
0 2,000 5,000 8,000 12,000 15,000
28 * 10-7 17 * 10-7 9.23 * 10-7 4.43 * 10-7 2.1 * 10-7
2,000 3,000 3,000 4,000 3,000
0.0165 0.0110 0.00591 0.00314 0.00137 0.00074
-0.0055 -0.0051 -0.00277 -0.00177 -0.00063
(b) The average rate of reaction is 1.05 * 10-6M>s. (c) From the slopes of the tangents to the graph, the rates are 12 * 10-7M>s at 5000 s, 5.8 * 10-7M>s at 8000 s. 14.23 (a) - ¢3H2O24>¢t = ¢3H24> ¢t = ¢3O24> ¢t (b) - 12 ¢3N2O4>¢t = 12 ¢3N24> ¢t = ¢3O24>¢t (c) - ¢3N24>¢t = -1>3¢3H24>¢t = -1>2¢3NH34>2¢t (d) - ¢3C2H5NH24>¢t = ¢3C2H44> ¢t = ¢3NH34> ¢t 14.25 (a) - ¢3O24>¢t = 0.24 mol>s; ¢3H2O4> ¢t = 0.48 mol>s (b) Ptotal decreases by 28 torr/min. 14.27 (a) If [A] doubles, there is no change in the rate or the rate constant. The overall rate is unchanged because [A] does not appear in the rate law; the rate constant changes only with a change in temperature. (b) The reaction is zero order in A, second order in B, and second order overall. (c) units of k = M-1 s-1 14.29 (a) Rate = k3N2O54 (b) Rate = 1.16 * 10-4 M>s (c) When the concentration of N2O5 doubles, the rate doubles. (d) When the concentration of N2O5 is halved, the rate doubles. 14.31 (a, b) k = 1.7 * 102 M-1s-1 (c) If 3OH-4 is tripled, the rate triples. (d) If 3OH-4 and 3CH3Br4 both triple, the rate increases by a factor of 9. 14.33 (a) Rate = k3OCl-43I -4 (b) k = 60 M-1 s-1 (c) Rate = 6.0 * 10-5 M>s 14.35 (a) Rate = k3BF343NH34 (b) The reaction is second order overall. (c) kavg = 3.41 M -1 s -1 (d) 0.170 M>s 14.37 (a) Rate = k3NO423Br24 (b) kavg = 1.2 * 104 M -2 s -1 (c) 12 ¢3NOBr4>¢t = - ¢3Br24>¢t (d) - ¢3Br24> ¢t = 8.4 M>s 14.39 (a) 3A40 is the molar concentration of reactant A at time zero. 3A4t is the molar concentration of reactant A at time t. t1>2 is the time required to reduce 3A40 by a factor of 2. k is the rate constant for a particular reaction. (b) A graph of ln[A] versus time yields a straight line for a first-order reaction. (c) On a graph of ln[A] versus time, the rate constant is the (–slope) of the straight line. 14.41 Plot [A] versus time. 14.43 (a) k = 3.0 * 10-6 s-1 (b) t1>2 = 3.2 * 104 s 14.45 (a) P = 30 torr (b) t = 51 s 14.47 Plot (ln PSO2Cl2) versus time, k = -slope = 2.19 * 10-5 s-1 14.49 (a) The plot of 1>3A4 versus time is linear, so the reaction is second order in [A]. (b) k = 0.040 M-1 min-1 (c) t1>2 = 38 min 14.51 (a) The plot of 1>3NO24 versus time is linear, so the reaction is second order in NO2. (b) k = slope = 10 M-1 s-1 (c) rate at 0.200 M = 0.400 M>s; rate at 0.100 M = 0.100 M>s; rate at 0.050 M = 0.025 M>s 14.53 (a) The energy of the collision and the orientation of the molecules when they collide determine whether a reaction will occur. (b) At a higher temperature, there are more total collisions and each collision is more
(b) Ea (reverse) = 73 kJ 14.59 (a) False. If you compare two reactions with similar collision factors, the one with the larger activation energy will be slower. (b) False. A reaction that has a small rate constant will have either a small frequency factor, a large activation energy, or both. (c) True. 14.61 Reaction (b) is fastest and reaction (c) is slowest. 14.63 (a) k = 1.1 s-1 (b) k = 13 s-1 (c) The method in parts (a) and (b) assumes that the collision model and thus the Arrhenious equation describe the kinetics of the reactions. That is, activation energy is constant over the temperature range under consideration. 14.65 A plot of ln k versus 1>T has a slope of -5.71 * 103; Ea = -R(slope) = 47.5 kJ>mol. 14.67 The reaction will occur 88 times faster at 50 °C, assuming equal initial concentrations. 14.69 (a) An elementary reaction is a process that occurs in a single event; the order is given by the coefficients in the balanced equation for the reaction. (b) A unimolecular elementary reaction involves only one reactant molecule; a bimolecular elementary reaction involves two reactant molecules. (c) A reaction mechanism is a series of elementary reactions that describes how an overall reaction occurs and explains the experimentally determined rate law. 14.71 A transition state is a high-energy complex formed when one or more reactants collide and distort in a way that can lead to formation of product(s). An intermediate is the product of an early elementary reaction in a multistep reaction mechanism. 14.73 (a) Unimolecular, rate = k3Cl 24 (b) bimo- lecular, rate = k3OCl-43H2O4 (c) bimolecular, rate = k3NO43Cl24 14.75 (a) Two intermediates, B and C. (b) three transition states (c) C ¡ D is fastest. (d) endothermic 14.77 (a) H2(g) + 2 ICl(g) ¡ I2(g) + 2 HCl(g) (b) HI is the intermediate. (c) If the first step is slow, the observed rate law is rate = k3H243ICl4. 14.79 The graph of 1/[NO] versus time is linear with positive slope, indicating that the reaction is second order in [NO]. The rate law obtained by assuming the second step is rate determining is rate = 3NO423Cl24. The two-step mechanism is consistent with the data. 14.81 (a) A catalyst is a substance that changes (usually increases) the speed of a chemical reaction without undergoing a permanent chemical change itself. (b) A homogeneous catalyst is in the same phase as the reactants, while a hetereogeneous catalyst is in a different phase. (c) A catalyst has no effect on the overall enthalpy change for a reaction, but it does affect activation energy. It can also affect the frequency factor. 14.83 (a) 270 Pt atoms in a 2.0-nm sphere (b) 200 Pt atoms on the surface of a 2.0-nm sphere (c) 74% Pt atoms on the surface (d) 4300 Pt atoms in a 5.0-nm sphere; 1300 Pt atoms on the surface; 30% Pt atoms on the surface (e) The 2-nm sphere will definitely be more catalytically active because it has a much greater percentage of its atoms on the surface where they can participate in the chemical reaction. 14.85 (a) Multiply the coefficients in the first reaction by 2 and sum. (b) NO2(g) is a catalyst because it is consumed and then reproduced in the reaction sequence. (c) This is a homogeneous catalysis. 14.87 (a) Use of chemically stable supports makes it possible to obtain very large surface areas per unit mass of the precious metal catalyst because the metal can be deposited in a very thin, even monomolecular, layer on the surface of the support. (b) The greater the surface area of the catalyst, the more reaction sites, the greater the rate of the catalyzed reaction. 14.89 To put
Answers to Selected Exercises two D atoms on a single carbon, it is necessary that one of the already existing C ¬ H bonds in ethylene be broken while the molecule is adsorbed, so that the H atom moves off as an adsorbed atom and is replaced by a D atom. This requires a larger activation energy than simply adsorbing C2H4 and adding one D atom to each carbon. 14.91 (a) Living organisms operate efficiently in a very narrow temperature range; the role of enzymes as homogeneous catalysts that speed up desirable reactions, without heating and undesirable side effects, is crucial for biological systems. (b) catalase: 2 H2O2 ¡ 2 H2O + O2; nitrogenase: N2 ¡ 2 NH3 (nitrogen fixation) (c) This model assumes that the rate of the bound substrate being chemically transformed into bound product is slow and rate determining. 14.93 Carbonic anyhdrase lowers the activation energy of the reaction by 42 kJ. 14.95 (a) The catalyzed reaction is approximately 10,000,000 times faster at 25 °C. (b) The catalyzed reaction is 180,000 times faster at 125 °C. 14.99 (a) Rate = 4.7 * 10-5 M>s (b, c) k = 0.84 M-2 s-1 (d) If the [NO] is increased by a factor of 1.8, the rate would increase by a factor of 3.2. 14.101 The reaction is second order in NO2. If 3NO240 = 0.100 M and 3NO24t = 0.025 M, use the integrated form of the second-order rate equation to solve for t. t = 48 s 14.105 (a) The half-life of 241Am is 4.3 * 102 yr, that of 125 I is 63 days (b) 125I decays at a much faster rate. (c) 0.13 mg of each isotope remains after 3 half-lives. (d) The amount of 241Am remaining after 4 days is 1.00 mg. The amount of 125I remaining after 4 days is 0.957 grams. 14.109 The plot of 1/[C5H6] versus time is linear and the reaction is second order. k = 0.167 M-1 s-1 14.112 (a) When the two elementary reactions are added, N2O2(g) appears on both sides and cancels, resulting in the overall reaction. 2NO(g) + H2(g) ¡ N2O(g) + H2O(g) (b) First reaction, -3NO4> ¢t = k3NO42; second reaction, -3H24>¢t = k3H243N2O24 (c) N2O2 is the intermediate. (d) Since [H2] appears in the rate law, the second step must be slow relative to the first. 14.115 (a) Cl2(g) + CHCl3(g) ¡ HCl(g) + CCl4(g) (b) Cl(g), CCl3(g) (c) reaction 1, unimolecular; reaction 2, bimolecular; reaction 3, bimolecular (d) Reaction 2 is rate determining. (e) Rate = k3CHCl343Cl241>2. 14.122 (a) k = 8 * 107 M -1 s -1 (b) N O O
N
F
(
O
N
F(
(c) NOF is bent with a bond angle of approximately 120°. (d) O N F
F
(e) The electron-deficient NO molecule is attracted to electron-rich F2, so the driving force for formation of the transition state is greater than simple random collisions.
CHAPTER 15 15.1 kf 7 kr (b) The equilibrium constant is greater than 1. 15.7 From the smallest to the largest equilibrium constant, (c) 6 (b) 6 (a). 15.11 Kc decreases as T increases, so the reaction is exothermic. 15.13 (a) Kp = Kc = 8.1 * 10-3. (b) Since kf 6 kr, in order for the two rates to be equal, [A] must be greater than [B], and the partial pressure of A is greater than the partial pressure of B. 15.15 (a) Kc = 3N2O43NO24>3NO43; homogeneous (b) Kc = 3CS243H244> 3CH443H2S42; homogeneous (c) Kc = 3CO44>3Ni(CO)44; Kc = 3H + 43F -4>3HF4; heterogeneous (d) homogeneous (e) Kc = 3Ag + 42> 3Zn2 + 4; heterogeneous (f) Kc = 3H + 4 3OH-4; homogeneous (g) Kc = 3H + 4 3OH-4; homogeneous 15.17 (a) Mostly reactants (b) mostly products 15.19 No, the equilibrium constant can never be a negative number. The equilibrium constant is a ratio of rate constants (or a ratio of concentrations), which are never negative. 15.21 Kp = 1.0 * 10-3 15.23 (a) The equilibrium favors NO and Br2 at this temperature. (b) Kc = 77 (c) Kc = 8.8 15.25 (a) Kp = 0.541 (b) Kp = 3.42 (c) Kc = 281 15.27 Kc = 0.14 15.29 Pure solids
A-19
and liquids are normally excluded from equilibrium-constant expressions because their concentrations, the ratio of moles of a substance to volume occupied by the substance, are constant. 15.31 (a) Kp = PO2 (b) Kc = 3Hg(solv)443O2(solv)4 15.33 Kc = 10.5 15.35 (a) Kp = 51 (b) Kc = 2.1 * 103 15.37 (a) 3H24 = 0.012 M, 3N24 = 0.019 M, 3H2O4 = 0.138 M (b) Kc = 653.7 = 7 * 102 15.39 (a) PCO2 = 4.10 atm, PH2 = 2.05 atm, PH2O = 3.28 atm (b) PCO2 = 3.87 atm, PH2 = 1.82 atm, PCO = 0.23 atm (c) Kp = 0.11 15.41 Kc = 2.0 * 104 15.43 (a) A reaction quotient is the result of a general set of concentrations whereas the equilibrium constant requires equilibrium concentrations. (b) to the right (c) The concentrations used to calculate Q must be equilibrium concentrations. 15.45 (a) Q = 1.1 * 10-8, the reaction will proceed to the left. (b) Q = 5.5 * 10-12, the reaction will proceed to the right. (c) Q = 2.19 * 10-10, the mixture is at equilibrium. 15.47 PCl2 = 5.0 atm 15.49 (a) 3Br24 = 0.00767 M, 3Br4 = 0.00282 M, 0.0451 g Br(g) (b) 3H24 = 0.014 M, 3I24 = 0.00859 M, 3Hl4 = 0.081 M, 21 g HI 15.51 3NO4 = 0.002 M, 3N24 = 3O24 = 0.087 M 15.53 The equilibrium pressure of Br2(g) is 0.416 atm. 15.55 (a) 3Ca2 + 4 = 3SO42-4 = 4.9 * 10-3 M (b) A bit more than 1.0 g CaSO4 is needed in order to have some undissolved CaSO4(s) in equilibrium with 1.4 L of saturated solution. 15.57 3IBr4 = 0.223 M, 3I24 = 3Br24 = 0.0133 M 15.59 (a) PCH3I = PHI = 0.422 torr, PCH4 = 104.7 torr, PI2 = 7.54 torr 15.61 (a) Shift equilibrium to the right (b) decrease the value of K (c) shift equilibrium to the left (d) no effect (e) no effect (f) shift equilibrium to the right 15.63 (a) No effect (b) no effect (c) no effect (d) increase equilibrium constant (e) no effect 15.65 (a) ¢H° = -155.7 kJ (b) The reaction is exothermic, so the equilibrium constant will decrease with increasing temperature. (c) ¢n does not equal zero, so a change in volume at constant temperature will affect the fraction of products in the equilibrium mixture. 15.67 An increase in pressure favors formation of ozone. 15.71 Kp = 24.7; Kc = 3.67 * 10-3 15.74 (a) PBr2 = 1.61 atm, PNO = 0.628 atm, PNOBr = 0.179 atm ; Kc = 0.0643 (b) Pt = 0.968 atm (c) 10.49 g NOBr 15.77 At equilibrium, PIBr = 0.21 atm, PI2 = PBr2 = 1.9 * 10-3 atm 15.80 Kp = 4.33, Kc = 0.0480 15.83 3CO24 = 3H24 = 0.264 M, 3CO4 = 3H2O4 = 0.236 M 15.87 (a) 26% of the CCl4 is converted to C and Cl2. (b) PCCl4 = 1.47 atm, PCl2 = 1.06 atm 15.91 Q = 8 * 10-6. Q 7 Kp, so the system is not at equilibrium; it will shift left to attain equilibrium. A catalyst that speeds up the reaction and thereby promotes the attainment of equilibrium would decrease the CO concentration in the exhaust. 15.93 At equilibrium, 3H6IO4-4 = 0.0015 M 15.97 At 850 °C, Kp = 14.1; at 950 °C, Kp = 73.8; at 1050 °C, Kp = 2.7 * 102; at 1200 °C, Kp = 1.7 * 103. Because K increases with increasing temperature, the reaction is endothermic.
CHAPTER 16 16.1 (a) HX, the H + donor, is the Brønsted–Lowry acid. NH3, the H + acceptor, is the Brønsted–Lowry base. (b) HX, the electron pair acceptor, is the Lewis acid. NH3, the electron pair donor, is the Lewis base. 16.3 (a) HY is a strong acid. There are no neutral HY molecules in solution, only H + cations and Y- anions. (b) HX has the smallest Ka value. It has most neutral acid molecules and fewest ions. (c) HX has fewest H + and highest pH. 16.5 (a) True. (b) False. Methyl orange turns yellow at a pH slightly greater than 4, so solution B could be at any pH greater than 4. (c) True. 16.7 (a) Molecule A, NH2OH (hydroxyl amine), acts as a base. Molecule A is an H + acceptor because of the nonbonded electron pair on the N atom of the amine ( ¬ NH2) group, not because it contains an ¬ OH group. (b) Molecule B, HCOOH (formic acid), acts as an acid. The molecule contains a ¬ COOH group where the H atom bonded to O is ionizable and HCOOH is an H + donor. (c) Molecule C, CH3OH (methanol), is an organic alcohol. The H atom bonded to O is not ionizable, and the ¬ OH group does not dissociate in aqueous solution; it is neither an acid nor a base. 16.9 (a) Molecule (b) is more acidic because its
A-20
Answers to Selected Exercises
conjugate base is resonance-stabilized and the ionization equilibrium favors the more stable products. (b) Increasing the electronegativity of X increases the strength of both acids. As X becomes more electronegative and attracts more electron density, the O ¬ H bond becomes weaker, more polar, and more likely to be ionized. An electronegative X group also stabilizes the anionic conjugate base, causing the ionization equilibrium to favor products and the value of Ka to increase. 16.11 Solutions of HCl and H2SO4 conduct electricity, taste sour, turn litmus paper red (are acidic), neutralize solutions of bases, and react with active metals to form H2(g). HCl and H2SO4 solutions have these properties in common because both compounds are strong acids. That is, they both ionize completely in H2O to form H + (aq) and an anion. (HSO4- is not completely ionized, but the first ionization step for H2SO4 is complete.) The presence of ions enables the solutions to conduct electricity; the presence of H + (aq) in excess of 1 * 10-7 M accounts for all other properties listed. 16.13 (a) The Arrhenius definition of an acid is confined to aqueous solution; the Brønsted–Lowry definition applies to any physical state. (b) HCl is the Brønsted–Lowry acid; NH3 is the Brønsted–Lowry base. 16.15 (a) (i) IO3- (ii) NH3 (b) (i) OH- (ii) H3PO4 16.17 Acid
ⴙ
Base Δ
Conjugate Acid
ⴙ
Conjugate Base
HCN(aq) NH3(aq) (a) NH4 + (aq) CN-(aq) + (b) H2O(l) (CH3)3 N(aq) (CH3)3NH (aq) OH-(aq) HPO42-(aq) HCOO-(aq) (c) HCOOH(aq) PO43-(aq)
16.19 (a) Acid: HC2O4-(aq) + H2O(l) Δ C2O42-(aq) + H3O + (aq); Base: HC2O4-(aq) + H2O(l) Δ H2C2O4(aq) + OH-(aq). (b) H2C2O4 is the conjugate acid of HC2O4- . C2O42- is the conjugate base of HC2O4-. 16.21 (a) CH3COO-, weak base; CH3COOH, weak acid (b) HCO3-, weak base; H2CO3, weak acid (c) O2-, strong base; OH-, strong base (d) Cl-, negligible base; HCl, strong acid (e) NH3, weak base; NH4 + , weak acid 16.23 (a) HBr. It is one of the seven strong acids. (b) F- . HCl is a stronger acid than HF, so F- is the stronger conjugate base. 16.25 (a) OH-(aq) + OH-(aq), the equilibrium lies to the right. (b) H2S(aq) + CH3COO-(aq), the equilibrium lies to the right. (c) HNO3(aq) + OH-(aq), the equilibrium lies to the left. 16.27 (a) No. In pure water, the only source of H + is the autoionization reaction, which produces equal concentrations of H + and OH- . As the temperature of water changes, the value of Kw changes, and the pH at which 3H + 4 = 3OH-4 changes. 16.29 (a) 3H + 4 = 2.2 * 10-11 M, basic (b) 3H + 4 = 1.1 * 10-6 M, acidic (c) 3H + 4 = 1.0 * 10-8 M, basic 16.31 3H + 4 = 3OH-4 = 3.5 * 10-8 M 16.33 (a) 3H + 4 changes by a factor of 100. (b) 3H + 4 changes by afactor of 3.2 16.35 (a) 3H + 4 decreases, pH increases (b) The pH is between 3 and 4. By calculation, pH = 3.2; the solution is acidic. (c) pH = 5.2 is between pH 5 and pH 6, closer to pH = 5. A good estimate is 7 * 10-6 M H + and 3 * 10-9 M OH- . By calculation, 3H + 4 = 6 * 10-6 M and 3OH-4 = 2 * 10-9 M. 16.37 [H + ] 7.5 2.8 5.6 5.0
* * * *
[OH -] -3
10 M 10-5 M 10-9 M 10-9 M
1.3 3.6 1.8 2.0
* * * *
-12
10 M 10-10 M 10-6 M 10-6 M
pH
pOH
Acidic or Basic
2.12 4.56 8.25 8.30
11.88 9.44 5.75 5.70
acidic acidic basic basic
16.39 3H + 4 = 4.0 * 10-8 M, 3OH-4 = 6.0 * 10-7 M, pOH = 6.22 16.41 (a) A strong acid is completely ionized in aqueous solution. (b) 3H + 4 = 0.500 M (c) HCl, HBr, HI 16.43 (a) 3H + 4 = 8.5 * 10-3 M, pH = 2.07 (b) 3H + 4 = 0.0419 M, pH = 1.377 (c) 3H + 4 = 0.0250 M, (d) pH = 1.602 3H + 4 = 0.167 M, pH = 0.778 16.45 (a) 3OH-4 = 3.0 * 10-3 M, pH = 11.48 (b) 3OH-4 = 0.3758 M, pH = 13.5750 (c) 3OH-4 = 8.75 * 10-5 M, pH = 9.942 (d) 3OH-4 = 0.17 M, pH = 13.23 16.47 3.2 * 10-3 M NaOH 16.49 (a) HBrO2(aq) Δ H + (aq) + BrO2-(aq), Ka = 3H + 43BrO2-4> 3HBrO24; HBrO2(aq) + H2O(l) Δ H3O + (aq) + BrO2-(aq), Ka = 3H3O + 43BrO2-4>3HBrO24 (b) C2H5COOH(aq) Δ H+(aq) + C2H5COO-(aq), Ka = 3H + 43C2H5COO-4> 3C2H5COOH4; C2H5COOH(aq) + H2O(l) Δ H3O + (aq) + C2H5COO-(aq), Ka = 3H3O + 43C2H5COO-4>3C2H5COOH4 16.51 Ka = 1.4 * 10-4 16.53 3H + 4 = 3ClCH2COO-4 = 0.0110 M, 3ClCH2COOH4 = 0.089 M , Ka = 1.4 * 10-3 16.55 0.089 M CH3COOH 16.57 3H + 4 = 3C6H5COO-4 = 1.8 * 10-3 M, 3C6H5COOH4 = 0.048 M 16.59 (a) 3H + 4 = 1.1 * 10-3 M , pH = 2.95 (b) 3H + 4 = 1.7 * 10-4 M, pH = 3.76 (c) 3OH-4 = 1.4 * 10-5 M, pH = 9.15 16.61 3H + 4 = 2.0 * 10-2 M, pH = 1.71 16.63 (a) 3H + 4 = 2.8 * 10-3 M, 0.69% ionization (b) 3H + 4 = 1.4 * 10-3 M, 1.4% ionization (c) 3H + 4 = 8.7 * 10-4 M, 2.2% ionization 16.65 HX(aq) Δ H + (aq) + X-(aq); Ka = 3H + 43X-4>3HX4. Assume that the percent of acid that ionizes is small. Let 3H + 4 = 3X-4 = y, Ka = y2>3HX4; y = K a1>23HX41>2. Percent ionization = y>3HX4 * 100. Substituting for y, percent ionization = 100 K a1>23HX41>2>3HX4 or 100 K a1>2>3HX41>2. That is, percent ionization varies inversely as the square root of the concentration of HX. 16.67 3H + 4 = 5.1 * 10-3M, pH = 2.29, 3C6H5O73-4 = 1.3 * 10-9 M. The approximation that the first ionization is less than 5% of the total acid concentration is not valid; the quadratic equation must be solved. The 3H + 4 produced from the second and third ionizations is small with respect to that present from the first step; the second and third ionizations can be neglected when calculating the 3H + 4 and pH. 3C6H5O73-4 is much less than 3H + 4. 16.69 (a) HONH3 + (b) When hydroxylamine acts as a base, the nitrogen atom accepts a proton. (c) In hydroxylamine, O and N are the atoms with nonbonding electron pairs; in the neutral molecule both have zero formal charges. Nitrogen is less electronegative than oxygen and more likely to share a lone pair of electrons with an incoming (and electron-deficient) H + . The resulting cation with the +1 formal charge on N is more stable than the one with the +1 formal charge on O. 16.71 (a) (CH3)2NH(aq) + H2O(l) Δ (CH3)2NH2 + (aq) + OH-(aq); Kb = 3(CH3)2NH2 + 43OH-4>3(CH3)2NH4 (b) CO32-(aq) + H2O(l) Δ HCO3-(aq) + OH-(aq) ; Kb = 3HCO3-43OH-4> 3(CO32-)4 (c) HCOO-(aq) + H2O(l) Δ HCOOH(aq) + OH-(aq) Kb = 3HCOOH43OH-4>3HCOO-4 16.73 From the quadratic formula, 3OH-4 = 6.6 * 10-3 M , pH = 11.82. 16.75 (a) 3C10H15ON4 = 0.033 M ,3C10H15ONH + 4 = 3OH-4 = 2.1 * 10-3 M (b) Kb = 1.4 * 10-4 16.77 (a) For a conjugate acid>conjugate base pair such as C6H5OH>C6H5O- , Kb for the conjugate base can always be calculated from Ka for the conjugate acid, so a separate list of Kb values is not necessary. (b) Kb = 7.7 * 10-5 (c) Phenolate is a stronger base than NH3. 16.79 (a) Acetic acid is stronger. (b) Hypochlorite ion is the stronger base. (c) For CH3COO-, Kb = 5.6 * 10-10; for ClO-, Kb = 3.3 * 10-7. 16.81 (a) 3OH-4 = 6.3 * 10-4 M, (b) (c) pH = 10.80 3OH-4 = 9.2 * 10-5 M, pH = 9.96 3OH-4 = 3.3 * 10-6 M, pH = 8.52 16.83 (a) Acidic (b) acidic (c) basic (d) neutral (e) acidic 16.85 Kb for the anion of the unknown salt is 1.4 * 10-11; Ka for the conjugate acid is 7.1 * 10-4. The conju-
Answers to Selected Exercises gate acid is HF and the salt is NaF. 16.87 (a) As the electronegativity of the central atom (X) increases, the strength of the oxyacid increases. (b) As the number of nonprotonated oxygen atoms in the molecule increases, the strength of the oxyacid increases. 16.89 (a) HNO3 is a stronger acid because it has one more nonprotonated oxygen atom and thus a higher oxidation number on N. (b) For binary hydrides, acid strength increases going down a family, so H2S is a stronger acid than H2O. (c) H2SO4 is a stronger acid because H + is much more tightly held by the anion HSO4- . (d) For oxyacids, the greater the electronegativity of the central atom, the stronger the acid, so H2SO4 is the stronger acid. (e) CCl3COOH is stronger because the electronegative Cl atoms withdraw electron density from other parts of the molecule, which weakens the O ¬ H bond and stabilizes the anionic conjugate base. Both effects favor increased ionization and acid strength. 16.91 (a) BrO- (b) BrO- (c) HPO42- 16.93 (a) True (b) False. In a series of acids that have the same central atom, acid strength increases with the number of nonprotonated oxygen atoms bonded to the central atom. (c) False. H2Te is a stronger acid than H2S because the H ¬ Te bond is longer, weaker, and more easily ionized than the H ¬ S bond. 16.95 Yes. The Arrhenius definition of a base, an OH-(aq) donor, is most restrictive; the Brønsted definition, an H+ acceptor, is more general; and the Lewis definition, an electron-pair donor, is most general. Any substance that fits the narrow Arrhenius definition will fit the broader Brønsted and Lewis definitions. 16.97 (a) Acid, Fe(ClO4)3 or Fe3 + ; base, H2O (b) Acid, H2O; base, CN- (c) Acid, BF3; base, (CH3)3N (d) Acid, HIO; base, NH2- 16.99 (a) Cu2 + , higher cation charge (b) Fe3 + , higher cation charge (c) Al3 + , smaller cation radius, same charge 16.101 (C2H5)3N is a stronger base than NH3 by virture of its smaller pKb. 16.104 K = 3.3 * 107 16.107 (a) For solutions with equal concentrations, the weaker acid will have a lower 3H + 4 and higher pH. (b) The acid with Ka = 8 * 10-5 is the weaker acid, so it has the higher pH. (c) The base with pKb = 4.5 is the stronger base, has the greater 3OH-4 and smaller 3H + 4, so higher pH. 16.109 Ka = 1.4 * 10-5 16.115 6.0 * 1013 H + ions 16.118 (a) To the precision of the reported data, the pH of rainwater 40 years ago was 5.4, no different from the pH today. With extra significant figures, 3H + 4 = 3.61 * 10-6 M, pH = 5.443 (b) A 20.0-L bucket of today’s rainwater contains 0.02 L (with extra significant figures, 0.0200 L) of dissolved CO2. 16.119 (a) Cl Al Cl Cl The electron-domain geometry and molecular structure are trigonal planar. (b) The Al atom is electron deficient. It acts like a Lewis acid in order to complete its octet. (c) Cl H Cl Cl Al Cl
NH3
Cl
Al
Cl
N
H H
+
(d) The Lewis theory is most appropriate. H and AlCl3 are both electron pair acceptors. 16.121 Rx 1, ¢H = 104 kJ; Rx 2, ¢H = -32 kJ. Reaction 2 is exothermic while reaction 1 is endothesmic. For binary acids with heavy atoms (X) in the same family, the longer and weaker the H ¬ X bond, the stronger the acid (and the more exothermic the ionization reaction). 16.124 (a) K(i) = 5.6 * 103, K(ii) = 10 (b) Both (i) and (ii) have K 7 1, so both could be written with a single arrow.
CHAPTER 17 17.1 The middle box has the highest pH. For equal amounts of acid HX, the greater the amount of conjugate base X- , the smaller the amount of H + and the higher the pH. 17.4 (a) Drawing 3 (b) Drawing 1 (c) Drawing 2 17.7 (a) The red curve corresponds to the more concentrated acid solution. (b) On the titration curve of a weak acid,
A-21
pH = pKa at the volume halfway to the equivalence point. At this volume, the red curve has the smaller pKa and the larger Ka. 17.10 (a) Q = 4.67 * 10-6; Q 6 Ksp and the solution is not saturat ed. (b) Ca(OH)2 precipitate forms in beaker (iii). 17.13 (a) The extent of ionization of a weak electrolyte is decreased when a strong electrolyte containing an ion in common with the weak electrolyte is added to it. (b) NaNO2 17.15 (a) 3H + 4 = 1.8 * 10-5 M, pH = 4.73 (b) 3OH-4 = 4.8 * 10-5 M, pH = 9.68 (c) 3H + 4 = 1.4 * 10-5 M, pH = 4.87 17.17 (a) 4.5% ionization (b) 0.018% ionization 17.19 In a mixture of CH3COOH and CH3COONa, CH3COOH reacts with added base and CH3COO- combines with added acid, leaving 3H + 4 relatively unchanged. Although HCl and Cl- are a conjugate acid–base pair, Cl- has no tendency to combine with added acid to form undissociated HCl. Any added acid simply increases 3H + 4 in an HCl ¬ NaCl mixture. 17.21 (a) pH = 3.82 (b) pH = 3.96 17.23 (a) pH = 5.26 (b) Na + (aq) + CH3COO-(aq) + H + (aq) + Cl-(aq) ¡ CH3COOH(aq) + Na + (aq) + Cl-(aq) + (c) CH3COOH(aq) + Na (aq) + OH-(aq) ¡ CH3COO-(aq) + H2O(l) + Na + (aq) 17.25 (a) pH = 1.58 (b) 36 g NaF 17.27 (a) pH = 4.86 (b) pH = 5.0 (c) pH = 4.71 17.29 (a) 3HCO3-4>3H2CO34 = 11 (b) 3HCO3-4>3H2CO34 = 5.4 17.31 360 mL of 0.10 M HCOONa, 640 mL of 0.10 M HCOOH 17.33 (a) Curve B (b) pH at the approximate equivalence point of curve A = 8.0, pH at the approximate equivalence point of curve B = 7.0 (c) For equal volumes of A and B, the concentration of acid B is greater, since it requires a larger volume of base to reach the equivalence point. 17.35 (a) False (b) true (c) true 17.37 (a) Above pH 7 (b) below pH 7 (c) at pH 7 17.39 The second color change of Thymol blue is in the correct pH range to show the equivalence point of the titration of a weak acid with a strong base. 17.41 (a) 42.4 mL NaOH soln (b) 35.0 mL NaOH soln (c) 29.8 mL NaOH soln 17.43 (a) pH = 1.54 (b) pH = 3.30 (c) pH = 7.00 (d) pH = 10.69 (e) pH = 12.74 17.45 (a) pH = 2.78 (b) pH = 4.74 (c) pH = 6.58 (d) pH = 8.81 (e) pH = 11.03 (f) pH = 12.42 17.47 (a) pH = 7.00 (b) 3HONH3 + 4 = 0.100 M, pH = 3.52 (c) 3C6H5 NH3 + 4 = 0.100 M, pH = 2.82 17.49 (a) The concentration of undissolved solid does not appear in the solubility product expression because it is constant. (b) Ksp = 3Ag + 43I -4; Ksp = 3Sr 2 + 43SO42-4; Ksp = 3Fe2 + 43OH-42; Ksp = 3Hg 22 + 43Br -42 17.51 (a) Ksp = 7.63 * 10-9 (b) Ksp = 2.7 * 10-9 (c) 5.3 * 10-4 mol Ba(IO3)2>L 17.53 Ksp = 2.3 * 10-9 17.55 (a) 7.1 * 10-7 mol AgBr>L (b) 1.7 * 10-11 mol AgBr>L (c) 5.0 * 10-12 mol AgBr>L 17.57 (a) The amount of CaF2(s) on the bottom of the beaker increases. (b) The 3Ca2 + 4 in solution increases. (c) The 3F -4 in solution decreases. 17.59 (a) 1.4 * 103 g Mn(OH)2>L (b) 0.014 g>L (c) 3.6 * 10-7 g>L 17.61 More soluble in acid: (a) ZnCO3 (b) ZnS (d) AgCN (e) Ba3(PO4)2 17.63 3Ni 2 + 4 = 1 * 10-8 M 17.65 (a) 9.1 * 10-9 mol AgI>L pure water (b) K = Ksp * Kf = 8 * 104 (c) 0.0500 mol AgI/L 0.100 M NaCN 17.67 (a) Q 6 Ksp; no Ca(OH)2 precipitates (b) Q 6 Ksp; no Ag2SO4 precipitates 17.69 pH = 11.5 17.71 AgI will precipitate first, at 3I-4 = 4.2 * 10-13 M. 17.73 AgCl will precipitate first. 17.75 The first two experiments eliminate group 1 and 2 ions (Figure 17.23). The absence of insoluble phosphate precipitates in the filtrate from the third experiment rules out group 4 ions. The ions that might be in the sample are those from group 3, Al3 + , Fe3 + , Cr3 + , Zn2 + , Ni2 + , Mn2 + , or Co2 + , and from group 5, NH4 + , Na + , or K + . 17.77 (a) Make the solution acidic with 0.2 M HCl; saturate with H2S. CdS will precipitate; ZnS will not. (b) Add excess base; Fe(OH3)(s) precipitates, but Cr3 + forms the soluble complex Cr(OH)4- . (c) Add (NH4)2HPO4; Mg 2 + precipitates as MgNH4PO4; K + remains soluble. (d) Add 6 M HCl; precipitate Ag + as AgCl(s); Mn2 + remains soluble. 17.79 (a) Base is required to increase 3PO43-4 so that the solubility product of the metal phosphates of interest is exceeded and the phosphate salts precipitate. (b) Ksp for the cations in group 3 is much larger, and so to exceed Ksp, a higher 3S2-4 is required. (c) They should all redissolve in strongly acidic solution.
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Answers to Selected Exercises
17.81 pOH = pKb + log5|BH-|>|B|6 17.83 (a) pH = 3.171 (b) pH = 2.938 (c) pH = 12.862 17.86 (a) pH of buffer A = pH of buffer B = 3.74. For buffers containing the same conjugate acid and base components, the pH is determined by the ratio of concentrations of conjugate acid and base. Buffers A and B have the same ratio of concentrations, so their pH values are equal. (b) Buffer capacity is determined by the absolute amount of buffer components available to absorb strong acid or strong base. Buffer A has the greater capacity because it contains the greater absolute concentrations of HCOOH and HCOO-. (c) Buffer A: pH = 3.74, ¢pH = 0.00; buffer B: pH = 3.66, ¢pH = -0.12 (d) Buffer A: pH = 3.74, ¢pH = 0.00; buffer B: pH = 2.74, ¢pH = -1.00 (e) The results of parts (c) and (d) are quantitative confirmation that buffer A has a significantly greater capacity than buffer B. 17.88 (a) molar mass = 82.2 g>mol (b) Ka = 3.8 * 10-7 17.90 At the halfway point, mol HA = mol Aand 3HA4 = 3A-4. 3conj. base4
3A-4
. If 3A-4>3HA4 = 1, 3conj. acid4 3HA-4 log(1) = 0 and pH = pKa of the weak acid being titrated. 17.92 pH = 7.75 17.93 1.6 L of 1.0 M NaOH 17.96 (a) CdS (b) BaCrO4 (c) NiCO3 (d) Ag2SO4 17.100 The solubility of Mg(OH)2 in 0.50 M NH4Cl is 0.11 mol> L 17.101 3KMnO44 = 3MnO4-4 = 0.11 M 17.104 3OH-4 = 1.7 * 10-11 M, pH of the buffer = 3.22 17.107 (a) The molar solubility of Cd(OH)2 is 1.8 * 10-5 mol>L. (b) The initial concentration of NaBr required to increase the molar solubility of Cd(OH)2 to 1.0 * 10-3 mol> L is 2 M. 17.108 (a) H + (aq) + HCOO-(aq) ¡ HCOOH(aq) (b) K = 5.6 * 103 (c) 3Na + 4 = 3Cl-4 = 0.075 M, 3H + 4 = 3HCOO-4 = 3.7 * 10-3 M, 3HCOOH4 = 0.071 M 17.114 3Sr 2 + 4 = 3SO42-4 = 5.7 * 10-4 M, Ksp = 3.2 * 10-7 pH = pKa + log
= pKa + log
CHAPTER 18 18.1 (a) A greater volume than 22.4 L (b) The gas will occupy more volume at 85 km than at 50 km. (c) We expect gases to behave most ideally in the thermosphere, around the stratopause and in the troposphere at low altitude. 18.6 Salt water contains high concentrations of dissolved salts and solids. It includes the world ocean (97.2% of all water, approximately 35,000 ppm of dissolved salts) and brackish or salty water (0.1% of all water). Freshwater (0.6% of all water on earth) refers to natural waters that have low concentrations (less than 500 ppm) of dissolved salts and solids. It includes the waters of lakes, rivers, ponds, and streams. Groundwater is freshwater that is under the soil. It resides in aquifers, porous rock that holds water, and composes 20% of the world’s freshwater. 18.9 The basic goals of green chemistry are to minimize or eliminate solvents and waste, generate nontoxic waste, be energy efficient, employ renewable starting materials, and take advantage of catalysts that enable the use of safe and common reagents. 18.11 (a) Its temperature profile (b) troposphere, 0 to 12 km; stratosphere, 12 to 50 km; mesosphere, 50 to 85 km; thermosphere, 85 to 110 km 18.13 (a) The partial pressure of O3 is 3.0 * 10-7 atm (2.2 * 10-4 torr). (b) 7.3 * 1015 O3 molecules/1.0 L air 18.15 8.6 * 1016 CO molecules/1.0 L air 18.17 (a) 570 nm (b) visible electromagnetic radiation 18.19 (a) Photodissociation is cleavage of a bond such that two neutral species are produced. Photoionization is absorption of a photon with sufficient energy to eject an electron, producing an ion and the ejected electron. (b) Photoionization of O2 requires 1205 kJ>mol. Photodissociation requires only 495 kJ/mol. At lower elevations, high-energy short-wavelength solar radiation has already been absorbed. Below 90 km, the increased concentration of O2 and the availability of longer-wavelength radiation cause the photodissociation process to dominate. 18.21 Ozone depletion reactions, which involve only O3, O2, or O (oxidation state = 0), do not involve a change in oxidation state for oxygen atoms. Reactions involving ClO and one of the oxygen species with a zero oxidation state do involve a change in the oxidation state of oxygen atoms. 18.23 (a) A chlorofluorocarbon is a compound that contains chlorine, fluorine, and
carbon, while a hydrofluorocarbon is a compound that contains hydrogen, fluorine, and carbon. An HFC contains hydrogen in place of the chlorine present in a CFC. (b) HFCs are potentially less harmful than CFCs because their photodissociation does not produce Cl atoms, which catalyze the destruction of ozone. 18.25 (a) The C ¬ F bond requires more energy for dissociation than the C ¬ Cl bond and is not readily cleaved by the available wavelengths of UV light. (b) Chlorine is present as chlorine atoms and chlorine oxide molecules, Cl and ClO, respectively. 18.27 (a) Methane, CH4, arises from decomposition of organic matter by certain microorganisms; it also escapes from underground gas deposits. (b) SO2 is released in volcanic gases and also is produced by bacterial action on decomposing vegetable and animal matter. (c) Nitric oxide, NO, results from oxidation of decomposing organic matter and is formed in lightning flashes. 18.29 (a) H2SO4(aq) + CaCO3(s) ¡ CaSO4(s) + H2O(l) + CO2(g) (b) The CaSO4(s) would be much less reactive with acidic solution, since it would require a strongly acidic solution to shift the relevant equilibrium to the right: CaSO4(s) + 2 H + (aq) Δ Ca2 + (aq) + 2 HSO4-(aq). CaSO4 would protect CaCO3 from attack by acid rain, but it would not provide the structural strength of limestone. 18.31 (a) Ultraviolet (b) 357 kJ>mol (c) The average C ¬ H bond energy from Table 8.4 is 413 kJ>mol. The C ¬ H bond energy in CH2O, 357 kJ>mol, is less than the “average” C ¬ H bond energy. (d) O O H
C
H hn
H
C H
18.33 Incoming and outgoing energies are in different regions of the electromagnetic spectrum. CO2 is transparent to incoming visible radiation but absorbs outgoing infrared radiation. 18.35 0.099 M Na + 18.37 (a) 3.22 * 103 g H2O (b) The final temperature is 43.4 °C. 18.39 4.361 * 105 g CaO 18.41 (a) Groundwater is freshwater (less than 500 ppm total salt content) that is under the soil; it composes 20% of the world’s freshwater. (b) An aquifer is a layer of porous rock that holds groundwater. 18.43 The minimum pressure required to initiate reverse osmosis is greater than 5.1 atm. 18.45 (a) CO2(g), HCO3-, H2O(l), SO42- , NO3- , HPO42- , H2PO4- (b) CH4(g), H2S(g), NH3(g), PH3(g) 18.47 25.1 g O2 18.49 Mg2 + (aq) + Ca(OH)2(s) ¡ Mg(OH)2(s) + Ca2 + (aq) 18.51 0.42 mol Ca(OH)2, 0.18 mol Na2CO3 18.53 4 FeSO4(aq) + O2(aq) + 2 H2O(l) ¡ 4 Fe 3 + (aq) + 4 OH-(aq) + 4 SO42-(aq); Fe 3 + (aq) + 3 HCO3-(aq) ¡ Fe(OH)3(s) + 3 CO2(g) 18.55 (a) Trihalomethanes are the byproducts of water chlorination; they contain one central carbon atom bound to one hydrogen and three halogen atoms. (b) H H Cl
C
Cl
Cl
C
Br
Cl Cl 18.57 The fewer steps in a process, the less waste is generated. Processes with fewer steps require less energy at the site of the process and for subsequent cleanup or disposal of waste. 18.59 (a) H2O (b) It is better to prevent waste than to treat it. Atom economy. Less hazardous chemical synthesis and inherently safer for accident prevention. Catalysis and design for energy efficiency. Raw materials should be renewable. 18.61 (a) Water as a solvent, by criteria 5, 7, and 12. (b) Reaction temperature of 500 K, by criteria 6, 12, and 1. (c) Sodium chloride as a by-product, according to criteria 1, 3, and 12. 18.66 Multiply Equation 18.7 by a factor of 2; then add it to Equation 18.9. 2 Cl(g) and 2 ClO(g) cancel from each side of the resulting equation to produce Equation 18.10. 18.69 Although HFCs have long lifetimes in the stratosphere, it is infrequent that light with energy sufficient to dissociate a C-F bond will reach an HFC molecule. F atoms, the bad actors in ozone destruction, are much less likely than Cl atoms to be produced by photodissociation in the stratosphere. 18.71 The formation of NO(g) is endothermic, so K increases with increasing
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Answers to Selected Exercises temperature. The oxidation of NO(g) to NO2(g) is exothermic, so the value of K decreases with increasing temperature. 18.75 7.1 * 108 m2 18.77 (a)CO32- is a relatively strong Brønsted–Lowry base and produces OH– in aqueous solution. If 3OH-(aq)4 is sufficient for the reaction quotient to exceed Ksp for Mg(OH)2, the solid will precipitate. (b) At these ion concentrations, Q 7 Ksp and Mg(OH)2 will precipitate. 18.81 (a) 2.5 * 107 ton CO2, 4.2 * 105 ton SO2 5 (b) 4.3 * 10 ton CaSO3 18.84 (a) H
O
H
H
O
H
(b) 258 nm (c) The overall reaction is O3(g) + O(g) ¡ 2 O2(g). OH(g) is the catalyst in the overall reaction because it is consumed and then reproduced. 18.86 The enthalpy change for the first step is -141 kJ, for the second step, -249 kJ, for the overall reaction, -390 kJ. 18.90 (a) (b) kavg = 1.13 * 1044 M-1 s-1 Rate = k3O343H4 18.95 (a) Process (i) is greener because it involves neither the toxic reactant phosgene nor the by-product HCl. (b) Reaction (i): C in CO2 is linear with sp hybridization; C in R ¬ N “ C “ O is linear with sp hybridization; C in the urethane monomer is trigonal planar with sp2 hybridization. Reaction (ii): C in COCl2 is linear with sp2 hybridization; C in R ¬ N “ C “ O is linear with sp hybridization; C in the urethane monomer is trigonal planar with sp2 hybridization. (c) The greenest way to promote formation of the isocyanate is to remove by-product, either water or HCl, from the reaction mixture.
CHAPTER 19 19.1 (a)
ature. ¢E is not zero for the process. 19.23 (a) At constant temperature, ¢S = qrev>T, where qrev is the heat that would be transferred if the process were reversible. (b) No. ¢S is a state function, so it is independent of path. 19.25 (a) Entropy increases. (b) 89.2 J/K 19.27 (a) For a spontaneous process, the entropy of the universe increases; for a reversible process, the entropy of the universe does not change. (b) For a reversible process, if the entropy of the system increases, the entropy of the surroundings must decrease by the same amount. (c) For a spontaneous process, the entropy of the universe must increase, so the entropy of the surroundings must decrease by less than 42 J/K. 19.29 (a) Positive ¢S (b) ¢S = 1.02 J>K (c) Temperature need not be specified to calculate ¢S, as long as the expansion is isothermal. 19.31 (a) Yes, the expansion is spontaneous. (b) As the ideal gas expands into the vacuum, there is nothing for it to “push back,” so no work is done. Mathematically, w = -Pext ¢V. Since the gas expands into a vacuum, Pext = 0 and w = 0. (c) The “driving force” for the expansion of the gas is the increase in entropy. 19.33 (a) An increase in temperature produces more available microstates for a system. (b) A decrease in volume produces fewer available microstates for a system. (c) Going from liquid to gas, the number of available microstates increases. 19.35 (a) ¢S is positive. (b) S of the system clearly increases in 19.11 (b) and (e); it clearly decreases in 19.9 (c). The entropy change is difficult to judge in 19.9 (a) and definition of the system in (d) is problematic. 19.37 S increases in (a) and (c); S decreases in (b). 19.39 (a) The entropy of a pure crystalline substance at absolute zero is zero. (b) In translational motion the entire molecule moves in a single direction; in rotational motion the molecule rotates or spins around a fixed axis. In vibrational motion the bonds within a molecule stretch and bend, but the average position of the atoms does not change. (c) H
H
translational Cl
H
rotational
H
Cl
Cl
H Cl vibrational
H
Cl
19.41 (a) Ar(g) (b) He(g) at 1.5 atm (c) 1 mol of Ne(g) in 15.0 L (d) CO2(g) 19.43 (a) ¢S 6 0 (b) ¢S 7 0 (c) ¢S 6 0 (d) ¢S L 0 19.45 (a)
Entrophy, S
(b) ¢H = 0 for mixing ideal gases. ¢S is positive because the disorder of the system increases. (c) The process is spontaneous and therefore irreversible. (d) Since ¢H = 0, the process does not affect the entropy of the surroundings. 19.4 ¢S is positive. 19.7 (a) At 300 K, ¢G = 0, and the system is at equilibrium. (b) The reaction is spontaneous at temperatures above 300 K. 19.10 (a) The minimum in the plot is the equilibrium position of the reaction. (b) The quantity x is ¢G°. 19.11 Spontaneous: a, b, c, d; nonspontaneous: e 19.13 (a) NH4NO3(s) dissolves in water, as in a chemical cold pack. Naphthalene (moth balls) sublimes at room temperature. (b) Melting of a solid is spontaneous above its melting point but nonspontaneous below its melting point. 19.15 (a) Endothermic (b) above 100 °C (c) below 100 °C (d) at 100 °C 19.17 (a) For a reversible process, the forward and reverse changes occur by the same path. In a reversible process, both the system and the surroundings are restored to their original condition by exactly reversing the change. A reversible change produces the maximum amount of work. (b) There is no net change in the surroundings. (c) The vaporization of water to steam is reversible if it occurs at the boiling temperature of water for a specified external (atmospheric) pressure and if the required heat is added infinitely slowly. (d) No. Natural processes are spontaneous in the direction they occur and nonspontaneous in the opposite direction. By definition they are irreversible. 19.19 (a) If the ideal gas is contained in a closed system at constant volume, a decrease in external temperature leads to a decrease in both temperature and pressure of the gas. (b) If the ideal gas is contained in a closed system at constant pressure, a decrease in external temperature leads to a decrease in both temperature and volume of the gas. (c) No. ¢E is a state function. ¢E = q + w; q and w are not state functions. Their values do depend on path, but their sum, ¢E, does not. 19.21 (a) An ice cube can melt reversibly at the conditions of temperature and pressure where the solid and liquid are in equilibrium. (b) We know that melting is a process that increases the energy of the system even though there is no change in temper-
Cl
0º C
100º C
(b) Boiling water, at 100 °C, has a much larger entropy change than melting ice at 0 °C. 19.47 (a) C2H6(g) (b) CO2(g) 19.49 (a) Sc(s), 34.6 J>mol-K; Sc(g), 174.7 J>mol-K. In general, the gas phase of a substance has a larger S° than the solid phase because of the greater volume and motional freedom of the molecules. (b) NH3(g), 192.5 J>mol-K; NH3(aq), 111.3 J>mol-K. Molecules in the gas phase have more motional freedom than molecules in solution. (c) 1 mol of
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Answers to Selected Exercises
P4(g), 280 J>K; 2 mol of P2(g), 2(218.1) = 436.2 J>K. More particles have a greater motional energy (more available microstates). (d) C (diamond), 2.43 J>mol-K; C (graphite), 5.69 J>mol-K. The internal entropy in graphite is greater because there is translational freedom among planar sheets of C atoms, while there is very little freedom within the covalent-network diamond lattice. 19.51 For elements with similar structures, the heavier the atoms, the lower the vibrational frequencies at a given temperature. This means that more vibrations can be accessed at a particular temperature, resulting in greater absolute entropy for the heavier elements. 19.53 (a) ¢S° = -120.5 J>K. ¢S° is negative because there are fewer moles of gas in the products. (b) ¢S° = +176.6 J>K. ¢S° is positive because there are more moles of gas in the products. (c) ¢S° = +152.39 J>K. ¢S° is positive because the product contains more total particles and more moles of gas. (d) ¢S° = +92.3 J>K. ¢S° is positive because there are more moles of gas in the products. 19.55 (a) ¢G = ¢H - T¢S (b) If ¢G is positive, the process is nonspontaneous, but the reverse process is spontaneous. (c) There is no relationship between ¢G and rate of reaction. 19.57 (a) Exothermic (b) ¢S° is negative; the reaction leads to a decrease in disorder. (c) ¢G° = -9.9 kJ (d) If all reactants and products are present in their standard states, the reaction is spontaneous in the forward direction at this temperature. 19.59 (a) ¢H° = -537.22 kJ, ¢S° = 13.7 J>K, ¢G° = -541.40 kJ, ¢G° = ¢H° - T¢S° = - 541.31 kJ (b) ¢H° = -106.7 kJ, ¢S° = -142.2 kJ, ¢G° = -64.0 kJ, ¢G° = ¢H° - T¢S° = -64.3 kJ (c) ¢H° = -508.3 kJ, ¢S° = -178 kJ, ¢G° = -465.8 kJ, ¢G° = ¢H° - T¢S° = -455.1 kJ. The discrepancy in ¢G° values is due to experimental uncertainties in the tabulated thermodynamic data. (d) ¢H° = -165.9 kJ, ¢S° = 1.4 kJ, ¢G° = -166.2 kJ, ¢G° = ¢H° - T¢S° = -166.3 kJ 19.61 (a) ¢G° = -140.0 kJ, spontaneous (b) ¢G ° = +104.70 kJ, nonspontaneous (c) ¢G ° = +146 kJ, nonspontaneous (d) ¢G° = -156.7 kJ, spontaneous 19.63 (a) 2 C8H18(l) + 25 O2(g) ¡ 16 CO2(g) + 18 H2O(l) (b) Because ¢S° is positive, ¢G° is more negative than ¢H°. 19.65 (a) The forward reaction is spontaneous at low temperatures but becomes nonspontaneous at higher temperatures. (b) The reaction is nonspontaneous in the forward direction at all temperatures. (c) The forward reaction is nonspontaneous at low temperatures but becomes spontaneous at higher temperatures. 19.67 ¢S 7 60.8 J>K 19.69 (a) T = 330 K (b) nonspontaneous 19.71 (a) ¢H° = 155.7 kJ, ¢S° = 171.4 kJ. Since ¢S° is positive, ¢G° becomes more negative with increasing temperature. (b) ¢G° = 19 kJ. The reaction is not spontaneous under standard conditions at 800 K (c) ¢G° = -15.7 kJ. The reaction is spontaneous under standard conditions at 1000 K. 19.73 (a) Tb = 79 °C (b) From the Handbook of Chemistry and Physics, 74th Edition, Tb = 80.1 °C. The values are remarkably close; the small difference is due to deviation from ideal behavior by C6H6(g) and experimental uncertainty in the boiling point measurement and the thermodynamic data. 19.75 (a) C2H2(g) + 52O2(g) ¡ 2 CO2(g) + H2O(l) (b) -1299.5 kJ of heat produced>mol C2H2 burned (c) wmax = -1235.1 kJ>mol C2H2 19.77 (a) ¢G becomes more negative. (b) ¢G becomes more positive. (c) ¢G becomes more positive. 19.79 (a) ¢G° = -5.40 kJ (b) ¢G = 0.30 kJ 19.81 (a) ¢G° = -16.77 kJ, K = 870 (b) ¢G° = 8.0 kJ, K = 0.039 (c) ¢G° = -497.9 kJ, K = 2 * 1087 19.83 ¢H° = 269.3 kJ, ¢S° = 0.1719 kJ>K (a) PCO2 = 6.0 * 10-39 atm (b) PCO2 = 1.6 * 10-4 atm 19.85 (a) HNO2(aq) Δ H + (aq) + NO2-(aq) (b) ¢G° = 19.1 kJ (c) ¢G = 0 at equilibrium (d) ¢G = -2.7 kJ 19.87 (a) The thermodynamic quantities T, E, and S are state functions. (b) The quantities q and w depend on the path taken. (c) There is only one reversible path between states. (d) ¢E = qrev + wmax, ¢S = qrev / T. 19.91 (a) 16 arrangements (b) 1 arrangement (c) The gas will spontaneously adopt the state with the most possible arrangements for the molecules, the state with maximum disorder. 19.96 (a) For all three compounds listed, there are fewer moles of gaseous products than reactants in the formation reaction, so we expect ¢Sf° to be negative. If ¢Gf° = ¢Hf° - T¢Sf° and ¢Sf° is negative, -T¢Sf° is positive and ¢Gf° is more positive than
¢Hf°. (b) In this reaction, there are more moles of gas in products, ¢Sf° is positive, -T¢Sf° is negative and ¢Gf° is more negative than ¢Hf°. 19.100 (a) K = 4 * 1015 (b) An increase in temperature will decrease the mole fraction of CH3COOH at equilibrium. Elevated temperatures must be used to increase the speed of the reaction. (c) K = 1 at 836 K or 563 °C. 19.104 (a) ¢G = 8.77 kJ (b) wmin = 8.77 kJ. In practice, a larger than minimum amount of work is required. 19.108 (a) Acetone, ° = 88.4 J>mol-K; dimethyl ether, ¢Svap ° = 86.6 J>mol-K; ¢Svap ° = 110 J>mol-K; octane, ¢Svap ° = 86.3 J>mol-K; pyriethanol, ¢Svap ° = 90.4 J>mol-K. Ethanol does not obey Trouton’s rule. dine, ¢Svap (b) Hydrogen bonding (in ethanol and other liquids) leads to more ordering in the liquid state and a greater than usual increase in entropy upon vaporization. Liquids that experience hydrogen bonding are probably exceptions to Trouton’s rule. (c) Owing to strong hydrogen bonding interactions, water probably does not obey Trouton’s ° = 109.0 J>mol-K. (d) ¢Hvap for C6H5Cl L 36 kJ>mol rule. ¢Svap 19.113 (a) For any given total pressure, the condition of equal moles of the two gases can be achieved at some temperature. For individual gas pressures of 1 atm and a total pressure of 2 atm, the mixture is at equilibrium at 328.5 K or 55.5 °C. (b) 333.0 K or 60 °C (c) 374.2 K or 101.2 °C (d) The reaction is endothermic, so an increase in the value of K as calculated in parts (a)–(c) should be accompanied by an increase in T.
CHAPTER 20 20.1 In a Brønsted–Lowry acid–base reaction, H + is transferred from the acid to the base. In a redox reaction, one or more electrons are transferred from the reductant to the oxidant. The greater the tendency of an acid to donate H + , the lesser the tendency of its conjugate base to accept H + . The stronger the acid, the weaker its conjugate base. Similarly, the greater the tendency of a reduced species to donate electrons, the lesser the tendency of the corresponding oxidized species to accept electrons. The stronger the reducing agent, the weaker the corresponding oxidizing agent. 20.4 (a) Add 1 M A2 + (aq) to the beaker with the A(s) electrode. Add 1 M B2 + (aq) to the beaker with the B(s) electrode. Add a salt bridge to enable the flow of ions from one compartment to the other. (b) The A electrode functions as the cathode. (c) Electrons flow through the external circuit from the anode to the ° = 1.00 V. 20.7 (a) The sign cathode, from B to A in this cell. (d) Ecell of ¢G° is positive. (b) The equilibrium constant is less than one. (c) No. An electrochemical cell based on this reaction cannot accomplish ° = 0.799 V. work on its surroundings. 20.9 (a) Line 1 (b) Ered = Ered 20.13 (a) Oxidation is the loss of electrons. (b) Electrons appear on the products’ side (right side). (c) The oxidant is the reactant that is reduced. (d) An oxidizing agent is the substance that promotes oxidation; it is the oxidant. 20.15 (a) True (b) false (c) true 20.17 (a) I, +5 to 0; C, +2 to +4 (b) Hg, +2 to 0; N, -2 to 0 (c) N, +5 to +2; S, -2 to 0 (d) Cl, +4 to +3; O, -1 to 0 20.19 (a) TiCl4(g) + 2 Mg(l) ¡ Ti(s) + 2 MgCl2(l) (b) Mg(l) is oxidized; TiCl4(g) is reduced. (c) Mg(l) is the reductant; TiCl4(g) is the oxidant. 20.21 (a) Sn2 + (aq) ¡ Sn4 + (aq) + 2e-, oxidation (b) TiO2 (s) + 4 H + (aq) + 2e - ¡ Ti2 + (aq) + 2 H2O(l), reduction (c) ClO3- (aq) + 6 H + (aq) + 6e - ¡ Cl-(aq) + 3 H2O(l), reduction (d) N2(g) + 8 H + (aq) + 6e- ¡ 2 NH4 + (aq), reduction (e) 4 OH-(aq) ¡ O2(g) + 2 H2O(l) + 4e -, oxidation (f) SO32-(aq) + 2 OH-(aq) ¡ SO42-(aq) + H2O(l) + 2e -, oxidation (g) N2(g) + 6 H2O(l) + 6e - ¡ 2 NH3(g) + 6 OH-(aq), reducation 20.23 (a) Cr2O72-(aq) + I -(aq) + 8 H + (aq) ¡ 2 Cr 3 + (aq) + IO3-(aq) + 4 H2O(l); oxidizing agent, Cr2O72- ; reducing agent, (b) I4 MnO4-(aq) + 5 CH3OH(aq) + 12 H + (aq) ¡ 4 Mn2 + (aq) + 5 HCO2H(aq) + 12 H2O(aq); oxidizing agent, reducing agent, MnO4- ; CH3OH (c) I2(s) + 5 OCl (aq) + H2O(l) ¡ 2 IO3-(aq) + 5 Cl-(aq) + 2 H + (aq); oxidizing agent, OCl-; reducing agent, l2 (d) As2O3(s) +
Answers to Selected Exercises 2 NO3-(aq) + 2 H2O(l) + 2 H + (aq) ¡ 2 H3AsO4(aq) + N2O3(aq); oxidizing agent, NO3-; reducing agent, As2O3 (e) 2 MnO4-(aq) + Br -(aq) + H2O(l) ¡ 2 MnO2(s) + BrO3-(aq) + 2 OH-(aq); oxidizing agent, MnO4-; reducing agent, Br- (f) Pb(OH)42-(aq) + ClO-(aq) ¡ PbO2(s) + Cl-(aq) + 2 OH-(aq) + H2O(l); oxidizing agent, ClO-; reducing agent, Pb(OH)4220.25 (a) The reaction Cu2 + (aq) + Zn(s) ¡ Cu(s) + Zn2 + (aq) is occurring in both figures. In Figure 20.3 the reactants are in contact, while in Figure 20.4 the oxidation half-reaction and reduction halfreaction are occurring in separate compartments. In Figure 20.3 the flow of electrons cannot be isolated or utilized; in Figure 20.4 electrical current is isolated and flows through the voltmeter. (b) Na + cations are drawn into the cathode compartment to maintain charge balance as Cu2 + ions are removed. 20.27 (a) Fe(s) is oxidized, Ag + (aq) is reduced. (b) Ag + (aq) + e - ¡ Ag(s); Fe(s) ¡ Fe2 + (aq) + 2e- (c) Fe(s) is the anode, Ag(s) is the cathode. (d) Fe(s) is negative; Ag(s) is positive. (e) Electrons flow from the Fe electrode (-) toward the Ag electrode (+). (f) Cations migrate toward the Ag(s) cathode; anions migrate toward the Fe(s) anode. 20.29 Electromotive force, emf, is the potential energy difference between an electron at the anode and an electron at the cathode of a voltaic cell. (b)One volt is the potential energy difference required to impart 1 J of energy to a charge of 1 coulomb. (c) Cell potential, Ecell, is the emf of an electrochemical cell. 20.31 (a) 2 H + (aq) + 2e- ¡ H2(g) (b) A standard hydrogen electrode, SHE, has components that are at standard conditions, 1 M H + (aq) and H2(g) at 1 atm. (c) The platinum foil in a SHE serves as an inert electron carrier and a solid reaction surface. 20.33 (a) A standard reduction potential is the relative potential of a reduction half° = 0 V 20.35 (a) reaction measured at standard conditions. (b) Ered Cr2 + (aq) ¡ Cr3 + (aq) + e- ; Tl3 + (aq) + 2e- ¡ Tl + (aq) ° = 0.78 V (b) Ered (c) Switch e
e
Voltmeter Inert (Pt) anode
NO3
Na
NO3
Cr2 Cr3 NO3
NO3
Inert (Pt) cathode
TI1 TI3
Movement of cations Movement of anions 20.37 (a) E° = 0.823 V (b) E° = 1.89 V (c) E° = 1.211 V (d) E ° = 0.62 V 20.39 (a) 3 Ag + (aq) + Cr(s) ¡ 3 Ag(s) + Cr 3+(aq), E° = 1.54 V (b) Two of the combinations have essentially equal E° values: 2 Ag + (aq) + Cu(s) ¡ 2 Ag(s) + Cu2 + (aq), E° = 0.462 V; 3 Ni2 + (aq) + 2 Cr(s) ¡ 3 Ni(s) + 2 Cr3 + (aq), E° = 0.46 V 20.41 (a) Anode, Sn(s); cathode, Cu(s). (b) The copper electrode gains mass as Cu is plated out, and the tin electrode loses mass as Sn is oxidized. (c) Cu2 + (aq) + Sn(s) ¡ Cu(s) + Sn2 + (aq). (d) E° = 0.473 V. 20.43 (a) Mg(s) (b) Ca(s) (c) H2(g) (d) BrO3-(aq) 20.45 (a) Cl2(aq), strong oxidant (b) MnO4-(aq), acidic, strong oxidant (c) Ba(s) strong reductant (d) Zn(s), reductant 20.47 (a) Cu2 + (aq) 6 O2(g) 6 Cr2O72-(aq) 6 Cl2(g) 6 H2O2(aq)
A-25
(b) H2O2(aq) 6 I-(aq) 6 Sn2 + (aq) 6 Zn(s) 6 Al(s) 20.49 Al and H2C2O4 20.51 (a) 2 Fe2 + (aq) + S2O62-(aq) + 4 H + (aq) ¡ 3+ 2 Fe (aq) + 2 H2SO3(aq); 2 Fe2 + (aq) + N2O(aq) + 2 H + (aq) ¡ 2 Fe 3 + (aq) + N2(g) + H2O(l); Fe 2 + (aq) + VO2+ (aq) + 2 H + (aq) ¡ Fe3 + (aq) + VO2 + (aq) + H2O(l) (b) E ° = -0.17 V, ¢G° = 33 kJ; E° = -2.54 V, ¢G ° = 4.90 * 102 kJ; E° = 0.23 V, ¢G° = -22 kJ (c) K = 1.8 * 10-6 = 10-6 ; K = 1.2 * 10-86 = 10-86 ; ° = -0.113 V K = 7.8 * 103 = 8 * 103 20.53 ¢G° = 21.8 kJ, Ecell 20.55 (a) E° = 0.16 V, K = 2.54 * 105 = 3 * 105 (b) E° = 0.277 V, K = 2.3 * 109 (c) E° = 0.45 V , K = 1.5 * 1075 = 1075 20.57 (a) K = 9.8 * 102 (b) K = 9.5 * 105 (c) K = 9.3 * 108 20.59 (a) w max = -130 kJ>mol Sn 20.61 (a) The Nernst equation is applicable when the components of an electrochemical cell are at nonstandard conditions. (b) Q = 1 (c) Q decreases and E increases 20.63 (a) E decreases (b) E decreases (c) E decreases (d) no effect 20.65 (a) (b) (c) E° = 0.48 V E = 0.53 V E = 0.46 V 20.67 (a) E° = 0.46 V (b) E = 0.37 V 20.69 (a) The compartment with 3Zn2 + 4 = 1.00 * 10-2 M is the anode. (b) E° = 0 (c) E = 0.0668 V (d) In the anode compartment 3Zn2 + 4 increases; in the cathode compartment 3Zn2 + 4 decreases 20.71 E° = 0.763 V, pH = 1.6 20.73 (a) The emf of a battery decreases as it is used. The concentrations of products increase and the concentrations of reactants decrease, causing Q to increase and Ecell to decrease. (b) A D-size battery contains more reactants than a AA, enabling the D to provide power for a longer time. 20.75 (a) 464 g PbO2 (b) 3.74 * 105 C of charge transferred 20.77 (a) The anode (b) E° = 0.50 V (c) The emf of the battery, 3.5 V, is exactly the standard cell potential calculated in part (b). (d) At ambient conditions, E L E°, so log Q L 1. Assuming that the value of E° is relatively constant with temperature, the value of the second term in the Nernst equation is approximately zero at 37 °C, and E L 3.5 V. 20.79 (a) The cell emf will have a smaller value. (b) NiMH batteries use an alloy such as ZrNi2 as the anode material. This eliminates the use and disposal problems associated with Cd, a toxic heavy metal. 20.81 The main advantage of a fuel cell is that fuel is continuously supplied, so that it can produce electrical current for a time limited only by the amount of available fuel. For the hydrogen-oxygen fuel cell, this is also a disadvantage because volatile and explosive hydrogen must be acquired and stored. Alkaline batteries are convenient, but they have a short lifetime, and the disposal of their zinc and manganese solids is more problematic than disposal of water produced by the hydrogenoxygen fuel cell. 20.83 (a) anode: Fe(s) ¡ Fe2 + (aq) + 2e- ; cathode: O2(g) + 4 H + (aq) + 4e - ¡ 2 H2O(l) (b) 2 Fe2 + (aq) + 3 H2O(l) + 3 H2O(l) ¡ Fe2O3 # 3 H2O(s) + 6 H + (aq) + 2e- ; O2(g) + 4 H + (aq) + 4e- ¡ 2 H2O(l) 20.85 (a) Mg is called a ° than the pipe metal “sacrificial anode” because it has a more negative Ered and is preferentially oxidized when the two are coupled. It is sacrificed to ° for Mg2 + is -2.37 V, more negative than most preserve the pipe. (b) Ered metals present in pipes, including Fe and Zn. 20.87 Under acidic conditions, air (O2) oxidation of Zn(s), 1.99 V; Fe(s), 1.67 V; and Cu(s), 0.893 V are all spontaneous. When the three metals are in contact, Zn will act as a sacrificial anode for both Fe and Cu, but after the Zn is depleted, Fe will be oxidized (corroded). 20.89 (a) Electrolysis is an electrochemical process driven by an outside energy source. (b) By definition, electrolysis reactions are nonspontaneous. (c) 2 Cl-(l) ¡ Cl2(g) + 2e(d) When an aqueous solution of NaCl undergoes electrolysis, sodium metal is not formed because H2O is preferentially reduced to form H2(g). 20.91 (a) 236 g Cr(s) (b) 2.51 A 20.93 (a) 4.0 * 105 g Li (b) The minimum voltage required to drive the electrolysis is + 4.41 V. 20.95 Gold is less active than copper and thus more difficult to oxidize. When crude copper is refined by electrolysis, Cu is oxidized from the crude anode, but any metallic gold present in the crude copper is not oxidized,so it accumulates near the anode, available for collection. 20.97(a) 2 Ni + (aq) ¡ Ni(s) + Ni2 + (aq) (b) 3 MnO42-(aq) + 4 H + (aq) ¡ 2 MnO4-(aq) + MnO2(s) + 2 H2O(l)
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Answers to Selected Exercises
(c) 3 H2SO3(aq) ¡ S(s) + 2 HSO4-(aq) + 2 H + (aq) + H2O(l) (d) Cl2(aq) + 2 OH-(aq) ¡ Cl-(aq) + ClO-(aq) + H2O(l) 20.100 (a) E° = 0.627 V, spontaneous (b) E° = -0.82 V, nonspontaneous (c) E° = 0.93 V, spontaneous (d) E° = 0.183 V, spontaneous 20.104 K = 1.6 * 106 20.107 The ship’s hull should be made negative. The ship, as a negatively charged “electrode,” becomes the site of reduction, rather than oxidation, in an electrolytic process. 20.110 3.0 * 104 kWh required 20.112 (a) H2 is being oxidized and N2 is being reduced. (b) K = 6.9 * 105 (c) E° = 0.05755 V 20.115 (a) E° = 0.028 V (b) cathode: Ag + (aq) + e- ¡ Ag(s); anode: Fe 2 + (aq) ¡ Fe 3 + (aq) + e - (c) ¢S° = 148.5 J. Since ¢S° is positive, ¢G° will become more negative and E° will become more positive as temperature is increased. 20.118 Ksp for AgSCN is 1.0 * 10-12.
CHAPTER 21 21.1 (a) 24Ne; outside; reduce neutron-to-proton ratio via b decay (b) 32 Cl; outside; increase neutron-to-proton ratio via positron emission or orbital electron capture (c) 108Sn; outside; increase neutron-to-proton ratio via positron emission or orbital electron capture (d) 216Po; outside; nuclei with Z Ú 84 usually decay via a emission. 21.4 (a) 7 min (b) 0.1 min-1 (c) 30% (3>10) of the sample remains after 12 min. 10 11 12 13 14 15 16 17 18 19 (d) 88 41Nb 21.5 (a) 5B, 5B; 6C, 6C; 7N, 7N; 8O, 8O, 8O; 9F 14 11 13 15 18 11 (b) 6C (c) 6C, 7N, 8O, 9F (d) 6C 21.7 (a) 25 protons, 30 neutrons (b) 80 protons, 121 neutrons (c) 19 protons, 20 neutrons 90 0 21.9 (a) 10n (b) 42He or a (c) 00g or g 21.11 (a) 90 37Rb ¡ 38Sr + -1e 72 0 72 76 76 (b) 34Se + -1e (orbital electron) ¡ 33As (c) 36Kr ¡ 35Br + 01e 222 4 211 211 0 (d) 226 88Ra ¡ 86Rn + 2He 21.13 (a) 82 Pb ¡ 83 Bi + -1 b 50 50 0 179 0 179 (b) (c) Mn Cr + e W + e ¡ 24 25 1 74 -1 ¡ 73 Ta 266 4 (d) 230 21.15 7 alpha emissions, 4 beta emisTh Ra + He ¡ 90 88 2 sions 21.17 (a) Positron emission (for low atomic numbers, positron emission is more common than electron capture) (b) beta emission (c) beta emission (d) beta emission 21.19 (a) Stable: 39 19K, 20 neutrons is a magic number (b) stable: 209 , 126 neutrons is a magic number Bi 83 65 (c) stable: 58 28Ni even proton, even neutron more likely to be stable; 28Ni 4 40 126 has high neutron-to-proton ratio 21.21 (a) 2He (c) 20Ca (e) 82Pb 21.23 The alpha particle, 42He, has a magic number of both protons and neutrons, while the proton is an odd proton, even neutron particle. Alpha is a very stable emitted particle, which makes alpha emission a favorable process. The proton is not a stable emitted particle, and its formation does not encourage proton emission as a process. 21.25 Protons and alpha particles are positively charged and must be moving very fast to overcome electrostatic forces that would repel them from the target nucleus. Neutrons are electrically neutral and not 10 1 259 repelled by the nucleus. 21.27 (a) 252 98Cf + 5B ¡ 3 0n + 103Lr 2 3 4 1 1 11 4 (b) 1H + 2He ¡ 2He + 1H (c) 1H + 5B ¡ 3 2He 122 0 59 0 59 (d) 122 53 I ¡ 54 Xe + -1e (e) 26Fe ¡ -1e + 27Co 238 4 241 1 21.29 (a) (b) 147N + 42He ¡ 92U + 2He ¡ 94Pu + 0n 17 1 56 4 60 0 (c) 21.31 (a) True. The O + H Fe + He Cu + e ¡ 8 1 26 2 29 -1 decay rate constant and half-life are inversely related. (b) False. If X is not radioactive, its half-life is essentially infinity. (c) True. Changes in the amount of A would be substantial and measurable over the 40-year time frame, while changes in the amount of X would be very small and difficult to detect. 21.33 When the watch is 50 years old, only 6% of the tritium remains. The dial will be dimmed by 94%. 21.35 The source must be replaced after 2.18 yr or 26.2 months; this corresponds to August 2012. 21.37 (a) 1.1 * 1011 alpha particles emitted in 5.0 min (b) 9.9 mCi 21.39 k = 1.21 * 10-4 yr-1; t = 4.3 * 103 yr 21.41 k = 5.46 * 10-10 yr-1; t = 3.0 * 109 yr 21.43 The energy released when one mole of Fe2O3 reacts is 8.515 * 103 J. The energy released when one mole of 42He is formed from protons and neutrons is
2.73 * 1012 J. This is 3 * 108 or 300 million times as much energy as the thermite reaction. 21.45 ¢m = 0.2414960 amu, ¢E = 3.604129 * 10-11 J> 27Al nucleus required, 8.044234 * 1013 J>100 g 27Al 21.47 (a) Nuclear mass: 2H, 2.013553 amu; 4He, 4.001505 amu; 6Li, 6.0134771 amu (b) nuclear binding energy: 2H, 3.564 * 10-13J; 4He, 4.5336 * 10-12 J; 6Li, 5.12602 * 10-12 J (c) binding energy/nucleon: 2H, 1.782 * 10-13 J/nucleon; 4He, 1.1334 * 10-12 J/nucleon; 6Li, 8.54337 * 10-13 J/nucleon. This trend in binding energy/nucleon agrees with the curve in Figure 21.12. The anomalously high calculated value for 4He is also apparent on the figure. 21.49 (a) 1.71 * 105 kg>d (b) 2.1 * 108 g 235U 21.51 (a) 59 Co; it has the largest binding energy per nucleon, and binding energy gives rise to mass defect. 21.53 (a) Nal is a good source of iodine because iodine is a large percentage of its mass; it is completely dissociated into ions in aqueous solution, and iodine in the form of I-(aq) is mobile and immediately available for biouptake. (b) A Geiger counter placed near the thyroid immediately after ingestion will register background, then gradually increase in signal until the concentration of iodine in the thyroid reaches a maximum. Over time, iodine-131 decays, and the signal decreases. (c) The radioactive iodine will decay to 0.01% of the original amount in approximately 82 days. 21.55 235U 21.57 The control rods in a nuclear reactor regulate the flux of neutrons to keep the reaction chain self-sustaining and also prevent the reactor core from overheating. They are composed of materials such as boron or cadmium that absorb neutrons. 21.59 (a) 2 2 3 1 239 1 133 98 1 1H + 1H ¡ 2He + 0n (b) 92U + 0n ¡ 51 Sb + 41Nb + 9 0n 11 21.61 (a) ¢m = 0.006627 g>mol; ¢E = 5.956 * 10 J = 5.956 * 108kJ>mol 11H (b) The extremely high temperature is required to overcome electrostatic charge repulsions between the nuclei so that they can come together to react. 21.63 (a) Boiling water reactor (b) fast breeder reactor (c) gas-cooled reactor 21.65 Hydrogen abstraction: RCOOH + # OH ¡ RCOO # + H2O; deprotonation: RCOOH + OH- ¡ RCOO- + H2O. Hydroxyl radical is more toxic to living systems because it produces other radicals when it reacts with molecules in the organism. Hydroxide ion, OH-, on the other hand, will be readily neutralized in the buffered cell environment. The acid–base reactions of OH- are usually much less disruptive to the organism than the chain of redox reactions initiated by # OH radical. 21.67 (a) 5.3 * 108 dis>s, 5.3 * 108 Bq (b) 6.1 * 102 mrad, 6.1 * 10-3 Gy (c) 5.8 * 103 mrem, 36 36 0 35 5.8 * 10-2 Sv 21.69 210 82Pb 21.71 (a) 17Cl : 18Ar + -1e (b) Cl and 37Cl both have an odd number of protons but an even number of neutrons. 36Cl has an odd number of protons and neutrons, so it is less 62 stable than the other two isotopes. 21.73 (a) 63Li + 56 28Ni : 31Ga 248 147 141 88 84 (b) 40 (c) Ca + Cm Sm + Xe Sr + Kr ¡ 62 ¡ 20 96 54 38 36 116 56 40 238 70 1 102 (d) 46Pd + 28Ni 20Ca + 92U ¡ 30Zn + 4 0n + 2 41Nb 21.77 The C ¬ OH bond of the acid and the O ¬ H bond of the alcohol break in this reaction. Initially, 18O is present in the C ¬ 18OH group of the alcohol. In order for 18O to end up in the ester, the 18 O ¬ H bond of the alcohol must break. This requires that the C ¬ OH bond in the acid also breaks. The unlabeled O from the acid ends up in the H2O product. 21.79 7Be, 8.612 * 10-13 J>nucleon; 9 Be, 1.035 * 10-12 J>nucleon; 10Be: 1.042 * 10-12 J>nucleon. The binding energies/nucleon for 9Be and 10Be are very similar; that for 10 Be is slightly higher. 21.85 1.4 * 104 kg C8H18
CHAPTER 22 22.1 (a) C2H4, the structure on the left, is the stable compound. Carbon can form strong multiple bonds to satisfy the octet rule, while silicon cannot. (b) The geometry about the central atoms in C2H4 is trigonal planar. 22.3 Molecules (b) and (d) will have the seesaw structure shown in the figure. 22.6 (c) Density, the ratio of mass to volume, increases going down the family; only this trend is consistent with the data in the figure. Other properties do not match the trend because (a) electronegativity and (b) first ionization energy both decrease rather than increase going down the family. Trends for both (d) X ¬ X single bond enthalpy and (e) electron affinity are
Answers to Selected Exercises somewhat erratic, with the trends decreasing from S to Po, and anomalous values for the properties of O, probably owing to its small covalent radius. 22.9 The compound on the left, with the strained three-membered ring, will be the most generally reactive. The larger the deviation from ideal bond angles, the more strain in the molecule and the more generally reactive it is. 22.11 Metals: (b) Sr, (c) Mn, (e) Na; nonmetals: (a) P, (d) Se, (f) Kr; metalloids: none. 22.13 (a) O (b) Br (c) Ba (d) O (e) Co (f) Br 22.15 (a) N is too small a central atom to fit five fluorine atoms, and it does not have available d orbitals, which can help accommodate more than eight electrons. (b) Si does not readily form p bonds, which are necessary to satisfy the octet rule for both atoms in the molecule. (c) As has a lower electronegativity than N; that is, it more readily gives up electrons to an acceptor and is more easily oxidized. 22.17 (a) NaOCH3(s) + H2O(l) ¡ NaOH(aq) + CH3OH(aq) (b) CuO(s) + 2 HNO3(aq) ¡ Cu(NO3)2(aq) + H2O(l) (c) WO3(s) + 3 H2(g) ¡ W(s) + 3 H2O(g) (d) 4 NH2OH(l) + O2(g) ¡ 6 H2O(l) + 2 N2(g) (e) Al4C3(s) + 12 H2O(l) ¡ 4 Al(OH)3(s) + 3 CH4(g) 22.19 (a) 11H, protium; 21H, deuterium; 31H, tritium (b) in order of decreasing natural abundance: protium 7 deuterium 7 tritium (c) Tritium is radioactive. (d) 31H ¡ 32He + -10e 22.21 Like other elements in group 1A, hydrogen has only one valence electron and its most common oxidation number is +1. 22.23 (a) Mg(s) + 2 H + (aq) ¡ Mg2 + (aq) + H2(g) 1100 °C
(b) C(s) + H2O(g) ¡ CO(g) + 3 H2(g) 1100 °C
(c) CH4(g) + H2O(g) ¡ CO(g) + 3 H2(g) 22.25 (a) NaH(s) + H2O(l) ¡ NaOH(aq) + H2(g) (b) Fe(s) + H2SO4(aq) ¡ Fe2 + (aq) + H2(g) + SO42-(aq) (c) H2(g) + Br2(g) ¡ 2 HBr(g) (d) 2 Na(l) + H2(g) ¡ 2 NaH(s) ¢
(e) PbO(s) + H2(g) ¡ Pb(s) + H2O(g) 22.27 (a) Ionic (b) molecular (c) metallic 22.29 Vehicle fuels produce energy via combustion reactions. The combustion of hydrogen is very exothermic and its only product, H2O, is a nonpollutant. 22.31 Xenon has a lower ionization energy than argon; because the valence electrons are not as strongly attracted to the nucleus, they are more readily promoted to a state in which the atom can form bonds with fluorine. Also, Xe is larger and can more easily accommodate an expanded octet of electrons. 22.33 (a) Ca(OBr)2, Br, +1 (b) HBrO3, Br, +5 (c) XeO3, Xe, +6 (d) ClO4-, Cl, +7 (e) HIO2, I, +3 (f) IF5; I, +5; F, -1 22.35 (a) iron(III) chlorate, Cl, +5 (b) chlorous acid, Cl, +3 (c) xenon hexafluoride, F, -1 (d) bromine pentafluoride; Br, + 5; F, -1 (e) xenon oxide tetrafluoride, F, -1 (f) iodic acid, I, +5 22.37 (a) van der Waals intermolecular attractive forces increase with increasing number of electrons in the atoms. (b) F2 reacts with water: F2(g) + H2O(l) ¡ 2 HF(g) + O2(g). That is, fluorine is too strong an oxidizing agent to exist in water. (c) HF has extensive hydrogen bonding. (d) Oxidizing power is related to electronegativity. Electronegativity and oxidizing power decrease in the ¢ order given. 22.39(a) 2 HgO(s) ¡ 2 Hg(l) + O2(g) ¢
(b) 2 Cu(NO3)2(s) ¡ 2 CuO(s) + 4 NO2(g) + O2(g) (c) PbS(s) + 4 O3(g) ¡ PbSO4(s) + 4 O2(g) (d) 2 ZnS(s) + 3 O2(g) ¡ 2 ZnO(s) + 2 SO2(g) (e) 2 K2O2(s) + 2 CO2(g) ¡ 2 K2CO3(s) + O2(g) hv
(f) 3 O2(g) ¡ 2 O3(g) 22.41 (a) acidic (b) acidic (c) amphoteric (d) basic 22.43 (a) H2SeO3, Se, +4 (b) KHSO3, S, +4 (c) H2Te, Te, -2 (d) CS2, S, -2 (e) CaSO4, S, +6 (f) CdS, S, -2 (g) ZnTe, Te, -2 22.45 (a) 2 Fe 3 + (aq) + H2S(aq) ¡ 2 Fe 2 + (aq) + S(s) + 2 H + (aq) (b) Br2(l) + H2S(aq) ¡ 2 Br-(aq) + S(s) + 2 H + (aq) (c) 2 MnO4-(aq) + 6 H + (aq) + 5 H2S(aq) ¡ 2 Mn2 + (aq) + 5 S(s) + 8 H2O(l) + (d) 2 NO3 (aq) + H2S(aq) + 2 H (aq) ¡ 2 NO2(aq) + S(s) + 2 H2O(l)
22.47 (a)
A-27
2
Se
O
O
O
Trigonal pyramidal (b)
S
S
Cl
Cl
Bent (free rotation around S–S bond) O
(c) O
S
Cl
O
H
Tetrahedral (around S) 22.49 (a) SO2(s) + H2O(l) Δ H2SO3(aq) Δ H + (aq) + HSO3 - (aq) (b) ZnS(s) + 2 HCl(aq) ¡ ZnCl2(aq) + H2S(g) (c) 8 SO32-(aq) + S8(s) ¡ 8 S2O32-(aq) (d) SO3(aq) + H2SO4(l) ¡ H2S2O7(l) 22.51 (a) NaNO2, +3 (b) NH3, -3 (c) N2O, +1 (d) NaCN, -3 (e) HNO3, + 5 (f) NO2, +4 (g) N2, 0 (h) BN, -3 22.53 (a) O N O H O N O H The molecule is bent around the central oxygen and nitrogen atoms; the four atoms need not be coplanar. The right-most form does not minimize formal charges and is less important in the actual bonding model. The oxidation state of N is +3. (b) N
N
N
N
N
N
N
N
N
The molecule is linear. The oxidation state of N is -1>3. (c)
H
H
H
N
N
H
H
The geometry is tetrahedral around the left nitrogen, trigonal pyramidal around the right. The oxidation state of N is -2. (d)
O O
N
O
The ion is trigonal planar; it has three equivalent resonance forms. The oxidation state of N is +5. 22.55 (a) Mg3N2(s) + 6 H2O(l) ¡ 3 Mg(OH)2(s) + 2 NH3(aq) (b) 2 NO(g) + O2(g) ¡ 2 NO2(g), redox reaction (c) N2O5(g) + H2O(l) ¡ 2 H + (aq) + 2 NO3-(aq) (d) NH3(aq) + H + (aq) ¡ NH4 + (aq) (e) N2H4(l) + O2(g) ¡ N2(g) + 2 H2O(g), redox reaction 22.57 (a) HNO2(aq) + H2O(l) ¡ NO3-(aq) + 2e (b) N2(g) + H2O(l) ¡ N2O(aq) + 2 H + (aq) + 2e22.59 (a) H3PO3, +3 (b) H4P2O7, +5 (c) SbCl3, +3 (d) Mg3As2, +5 (e) P2O5, +5 (f) Na3PO4, +5 22.61 (a) Phosphorus is a larger atom than nitrogen, and P has energetically available 3d orbitals, which participate in the bonding, but nitrogen does not. (b) Only one of the three hydrogens in H3PO2 is bonded to oxygen. The other two are bonded directly to phosphorus and are not easily ionized. (c) PH3 is a weaker base
A-28
Answers to Selected Exercises
than H2O so any attempt to add H + to PH3 in the presence of H2O causes protonation of H2O. (d) The P4 molecules in white phosphorus have more severely strained bond angles than the chains in red phosphorus, causing white phosphorus to be more reactive. 22.63 (a) 2 Ca3PO4(s) + 6 SiO2(s) + 10 C(s) ¡ P4(g) + 6 CaSiO3(l) +10 CO(g) (b) PBr3(l) + 3 H2O(l) ¡ H3PO3(aq) + 3 HBr(aq) (c) 4 PBr3(g) + 6 H2(g) ¡ P4(g) + 12 HBr(g) 22.65 (a) HCN (b) Ni(CO)4 (c) Ba(HCO3)2 (d) CaC2 (e) K2CO3 ¢ 22.67 (a) ZnCO3(s) ¡ ZnO(s) + CO2(g) (b) BaC2(s) + 2 H2O(l) ¡ Ba2 + (aq) + 2 OH-(aq) + C2H2(g) (c) 2 C2H2(g) + 5 O2(g) ¡ 4 CO2(g) + 2 H2O(g) (d) CS2(g) + 3 O2(g) ¡ CO2(g) + 2 SO2(g) (e) Ca(CN)2(s) + 2 HBr(aq) ¡ CaBr2(aq) + 2 HCN(aq) 22.69 800 °C (a) 2 CH4(g) + 2 NH3(g) + 3 O2(g) ¡ 2 HCN(g) + 6 H2O(g) + (b) NaHCO3(s) + H (aq) ¡ CO2(g) + H2O(l) + Na + (aq) (c) 2 BaCO3(s) + O2(g) + 2 SO2(g) ¡ 2 BaSO4(s) + 2 CO2(g) 22.71 (a) H3BO3, +3 (b) SiBr4, +4 (c) PbCl2, +2 (d) Na2B4O7 # 10 H2O, +3 (e) B2O3, +3 (f) GeO2, +4 22.73 (a) Tin (b) carbon, silicon, and germanium (c) silicon 22.75 (a) Tetrahedral (b) Metasilicic acid will probably adopt the single-strand silicate chain structure shown in Figure 22.34 (b). The Si to O ratio is correct and there are two terminal O atoms per Si that can accommodate the two H atoms associated with each Si atom of the acid. 22.77 (a) Diborane has bridging H atoms linking the two B atoms. The structure of ethane has the C atoms bound directly, with no bridging atoms. (b) B2H6 is an electron-deficient molecule. The 6 valence electron pairs are all involved in B ¬ H sigma bonding, so the only way to satisfy the octet rule at B is to have the bridging H atoms shown in Figure 22.36. (c) The term hydridic indicates that the H atoms in B2H6 have more than the usual amount of electron density for a covalently bound H atom. 22.81 (a) SO2(g) + H2O(l) Δ H2SO3(aq) (b) Cl2O7(g) + H2O(l) Δ 2 HClO4(aq) (c) Na2O2(s) + 2 H2O(l) ¡ H2O2(aq) + 2 NaOH(aq) (d) BaC2(s) + 2 H2O(l) ¡ Ba2 + (aq) + 2 OH-(aq) + C2H2(g) (e) 2 RbO2(s) + 2 H2O(l) ¡ 2 Rb+(aq) + 2 OH-(aq) + O2(g) + H2O2(aq) (f) Mg3N2(s) + 6 H2O(l) ¡ 3 Mg(OH)2(s) + 2 NH3(g) (g) NaH(s) + H2O(l) ¡ NaOH(aq) + H2(g) 22.85 (a) PO43- , +5; NO3- , +5, (b) The Lewis structure for NO43- would be: 3
O O
N
O
O The formal charge on N is +1 and on each O atom is -1. The four electronegative oxygen atoms withdraw electron density, leaving the nitrogen deficient. Since N can form a maximum of four bonds, it cannot form a p bond with one or more of the O atoms to regain electron density, as the P atom in PO43- does. Also, the short N ¬ O distance would lead to a tight tetrahedron of O atoms subject to steric repulsion. 22.89 (a) 1.94 * 103 g H2 (b) 2.16 * 104 L H2 (c) 2.76 * 105 kJ 22.91 (a) -285.83 kJ/mol H2; -890.4 kJ/ mol CH4 (b) -141.79 kJ/g H2; -55.50 kJ/g CH4 (c) 1.276 * 104 kJ>m3 H2; 3.975 * 104 kJ>m3 CH4 22.95 (a) SO2(g) + 2 H2S(aq) ¡ 3 S(s) + 2 H2O(l) or 8 SO2(g) + (b) 16 H2S(aq) ¡ 3 S8(s) + 16 H2O(l) 4.0 * 103 mol = 9.7 * 104 L H2S (c) 1.9 * 105 g S produced 22.97 The average bond enthalpies are H ¬ O, 463 kJ; H ¬ S, 367 kJ; H ¬ Se, 316 kJ; H ¬ Te, 266 kJ. The H ¬ X bond enthalpy decreases steadily in the series. The origin of this effect is probably the increasing size of the orbital from X with which the hydrogen 1s orbital must overlap. 22.101 Dimethylhydrazine produces 0.0369 mol gas per gram of reactants, while methylhydrazine produces 0.0388 mol gas per gram of reactants. Methylhydrazine has marginally greater thrust.
22.103 (a) 3 B2H6(g) + 6 NH3(g) ¡ 2 (BH)3(NH)3(l) + 12 H2(g); 3 LiBH4(s) + 3 NH4Cl(s) ¡ 2 (BH)3(NH)3(l) + 9 H2(g) + 3 LiCl(s) (b) H H
B N
N B
H
H
H B N
H
H
H
H
B N
N B H
H
B N
H
H
H
H
B N
N B
B N
H H
H
(c) 2.40 g (BH)3(NH)3
CHAPTER 23 23.2 (a)
N
Cl Pt
N
Cl
(b) Coordination number = 4, coordination geometry = square planar (c) oxidation state = +2 23.4 aminotrichloroplatinate(II) 23.6 Molecules (1), (3), and (4) are chiral because their mirror images are not superimposible on the original molecules. 23.8 (a) diagram (4) (b) diagram (1) (c) diagram (3) (d) diagram (2) 23.11 The lanthanide contraction is the name given to the decrease in atomic size due to the build-up in effective nuclear charge as we move through the lanthanides (elements 58–71) and beyond them. This effect offsets the expected increase in atomic size, decrease in ionization energy, and increase in electron affinity going from period 5 to period 6 transition elements. This causes the chemical properties of period 5 and period 6 elements in the same family to be even more similar than we would expect. 23.13 (a) All transition metal atoms have two s-electrons in their valence shell. Loss of these s-electrons leads to the +2 oxidation state common for most of the transition metals. 23.15 (a) Ti3 + , 3Ar43d1 (b) Ru2 + , 3Kr44d6 (c) Au3 + , 3Xe44f145d8 (d) Mn4 + , 3Ar43d3 23.17 (a) The unpaired electrons in a paramagnetic material cause it to be weakly attracted into a magnetic field. A diamagnetic material, where all electrons are paired, is very weakly repelled by a magnetic field. 23.19 The diagram shows a material with misaligned spins that become aligned in the direction of an applied magnetic field. This is a paramagnetic material. 23.21 (a) In Werner’s theory, primary valence is the charge of the metal cation at the center of the complex. Secondary valence is the number of atoms bound or coordinated to the central metal ion. The modern terms for these concepts are oxidation state and coordination number, respectively. (b) Ligands are the Lewis base in metal–ligand interactions. As such, they must possess at least one unshared electron pair. NH3 has an unshared electron pair but BH3, with less than 8 electrons about B, has no unshared electron pair and cannot act as a ligand. 23.23 (a) +2 (b) 6 (c) 2 mol AgBr(s) will precipitate per mole of complex. 23.25 (a) Coordination number = 4, oxidation number = +2; 4 Cl- (b) 5, +4; 4 Cl- , 1 O2- (c) 6, +3; 4 N, 2 Cl- (d) 5, +2; 5 C (e) 6, +3; 6 O (f) 4, +2; 4 N 23.27 (a) A monodentate ligand binds to a metal via one atom, a bidentate ligand binds through two atoms. (b) Three bidentate ligands fill the coordination sphere of a six-coordinate complex. (c) A tridentate ligand has at least three atoms with unshared electron pairs in the correct orientation to simultaneously bind one or more metal ions. 23.29 (a) Ortho-phenanthroline, o-phen, is bidentate (b) oxalate, C2O42- , is bidentate (c) ethylenediaminetetraacetate, EDTA, is pentadentate (d) ethylenediamine, en, is bidentate. 23.31 (a) The term chelate effect refers to the special stability associated with formation of a metal complex containing a polydentate (chelate) ligand relative to a complex containing only monodentate ligands. (b) The increase in entropy, + ¢S, associated with the substitution of a chelating ligand for two or more monodentate ligands generally gives rise to the chelate effect. Chemical reactions with + ¢S tend to be spontaneous, have negative ¢G and large values of K. (c) Polydentate ligands are used as sequestering agents to bind metal ions and prevent them from under-
Answers to Selected Exercises going unwanted chemical reactions without removing them from solution. 23.33 The ligand is not typically a chelate. The entire molecule is planar and the benzene rings on either side of the two N atoms inhibit their approach in the correct orientation for chelation. 23.35 (a) 3Cr(NH3)64(NO3)3 (b) 3Co(NH3)4CO342SO4 (c) 3Pt(en)2Cl24Br2 (d) K3V(H2O)2Br44 (e) 3Zn(en)243Hgl44 23.37 (a) tetraamminedichlororhodium(III) chloride (b) potassium hexachlorotitanate(IV) (c) tetrachlorooxomolybdenum(VI) (d) tetraaqua(oxalato)platinum (IV) bromide 23.39 ONO NH3 ONO ONO Pd
Pd
(a) H3N
H3N
NH3 cis
ONO trans
(b) 3Pd(NH3)2(ONO)24, 3Pd(NH3)2(NO2)24
(c) N
N
N
Cl
N
N
Cl
N
Cl
xy
(f) d8
d3
high spin 23.67 3Pt(NH3)64Cl4; 3Pt(NH3)4Cl24Cl2; 3Pt(NH3)3Cl34Cl; 3Pt(NH3)2Cl44; K3Pt(NH3)Cl54 23.71 (a) H H H H H
C
P
C
C
P
C
H H
C
H H
H H
C
H H
H
(d) 3Co(NH3)4Br24Cl, 3Co(NH3)4BrCl4Br 23.41 Yes. No structural or stereoisomers are possible for a tetrahedral complex of the form MA2B2. The complex must be square planar with cis and trans geometric isomers. 23.43 (a) One isomer (b) trans and cis isomers with 180° and 90° Cl ¬ Ir ¬ Cl angles, respectively (c) trans and cis isomers with 180° and 90° Cl ¬ Fe ¬ Cl angles, respectively. The cis isomer is optically active. 23.45 (a) We cannot see the light with 300 nm wavelength, but we can see the 500 nm light. (b) Complementary colors are opposite each other on an artist’s color wheel. (c) A colored metal complex absorbs visible light of its complementary color. (d) 196 kJ>mol 23.47 No. All 6 d-electrons in a low-spin octahedral Fe(II) complex will pair and occupy the low-energy dxy, dxz, and dyz orbitals. With no unpaired electrons, the complex cannot be paramagnetic. 23.49 Most of the attraction between a metal ion and a ligand is electrostatic. Whether the interaction is ion–ion or ion–dipole, the ligand is strongly attracted to the metal center and can be modeled as a point negative charge. 23.51 (a) d 2 2, d 2
H
H
Both dmpe and en are bidentate ligands, binding through P and N, respectively. Because phosphorus is less electronegative than N, dmpe is a stronger electron pair donor and Lewis base than en. Dmpe creates a stronger ligand field and is higher on the spectrochemical series. Structurally, dmpe occupies a larger volume than en. M–P bonds are longer than M ¬ N bonds and the two -CH3 groups on each P atom in dmpe create more steric hindrance than the H atoms on N in en. μ (b) The oxidation state of Mo is zero. (c)The symbol P P represents the bidentate dmpe ligand. CN CO P CO P CN Mo Mo P P CO CN CN CO
P
CN
CN
NC
Mo P
CO
CN
P
Mo CO
OC
CO
P
z
optical isomers
dxy, dxz, dyz (b) The magnitude of ¢ and the energy of the d-d transition for a d1 complex are equal. (c) ¢ = 220 kJ>mol 23.53 A yellow color is due to absorption of light around 400 to 430 nm, a blue color to absorption near 620 nm. The shorter wavelength corresponds to a higher-energy electron transition and larger ¢ value. Cyanide is a stronger-field ligand, and its complexes are expected to have larger ¢ values than aqua complexes. 23.55 (a) Ti3 + , d1 (b) Co3 + , d6 (c) Ru3 + , d5 (d) Mo5 + , d1, (e) Re3 + , d4 23.57 Yes. A weak-field ligand leads to a small ¢ value and a small d-orbital splitting energy. If the splitting energy of a complex is smaller than the energy required to pair electrons in an orbital, the complex is high spin. 23.59 (a) Mn, 3Ar44s23d5; Mn2 + , 3Ar43d5; 1 unpaired electron (b) Ru, 3Kr45s14d7; Ru2 + , 3Kr44d6; 0 unpaired electrons (c) Rh, 3Kr45s14d8; Rh2 + , 3Kr44d7; 1 unpaired electron 23.61 All complexes in this exercise are six-coordinate octahedral. (a) (b) (c) d 4, high spin
d 5, low spin 23.63
N
V Cl
(e)
V N
(d)
A-29
d 5, high spin
d 6, low spin
23.74 (a) Hemoglobin is the iron-containing protein that transports O2 in human blood. (b) Chlorophylls are magnesium-containing porphyrins in plants. They are the key components in the conversion of solar energy into chemical energy that can be used by living organisms. (c) Siderophores are iron-binding compounds or ligands produced by a microorganism. They compete on a molecular level for iron in the medium outside the organism and carry needed iron into the cells of the organism. 23.76 (a) Pentacarbonyliron(0) (b) The oxidation state of iron must be zero. (c) Two. One isomer has CN in an axial position and the other has it in an equatorial position. 23.78 (a)
d2 (b) Visible light with l = hc>¢ is absorbed by the complex, promoting one of the d electrons into a higher-energy d-orbital. The remaining wavelengths are reflected or transmitted; the combination of these wavelengths is the color we see. (c) 3V(H2O)643 + will absorb light with
A-30
Answers to Selected Exercises
higher energy because it has a larger ¢ than 3VF643- . H2O is in the middle of the spectrochemical series and causes a larger ¢ than F- , a weak-field ligand. 23.80 3Co(NH3)643 + , yellow; 3Co(H2O)642 + , pink; 3CoCl442- , blue 23.85 (a) NC
O C
2
C O
H2C
CN
CHAPTER 24 24.1 Molecules (c) and (d) are the same molecule. 24.4 Compound (b), which has hydrogen bonding, has the highest boiling point. 24.7 (a) sp3 (b) sp2 (c) sp2 (d) sp 24.9 Numbering from the right on the condensed structural formula, C1 has trigonal-planar electrondomain geometry, 120° bond angles, and sp2 hybridization; C2 and C5 have tetrahedral electron-domain geometry, 109° bond angles, and sp3 hybridization; C3 and C4 have linear electron-domain geometry, 180° bond angles, and sp hybridization. 24.11 NH3 and CO are not typical organic molecules. NH3 contains no carbon atoms. Carbon monoxide contains a C atom that does not form four bonds. 24.13 (a) A straight-chain alkane has all carbon atoms connected in a continuous chain. A carbon atom is bound to no more than two other carbon atoms and forms only s bonds. A branched-chain hydrocarbon has a branch; at least one carbon atom is bound to three or more carbon atoms. (b) An alkane is a complete molecule composed of carbon and hydrogen in which all bonds are s bonds. An alkyl group is a substituent formed by removing a hydrogen atom from an alkane. 24.15 (a) 2-methylhexane (b) 4-ethyl-2,4-dimethyldecane (c) CH3CH2CH2CH2CH2CH(CH3)2
H2C
H H H
C
C
C
C C
H
CH2
CH3
C
H
CH2
H CH3 H
or
CH3
H2C
CH
H2C
CH CH2
24.17 (a) 2,3-dimethylheptane (b) CH3CH2CH2C(CH3)3
CH3
CH
CH3CH2CH2CH2C H
H
CH3C
C
CCH3
CH3CH2CH2C H
CH2CH
CH2
CH3C
C
CH2CH
CH2
H CH3
CH HC
CH2
H 2C
CH2
CH2 24.29 (a)
CH
CH
C CH2
H 2C
CH C
H2C
CH2
H
CH2
CH3 H
CH3 C
C
CH3CH2
H CH3
(b) CH3CH2CH2
C
CH3 CH
CH2
CH
CH3
(c) cis-6-methyl-3-octene (d) para-dibromobenzene (e) 4,4-dimethyl1-hexyne 24.31 Geometric isomerism in alkenes is the result of restricted rotation about the double bond. In alkanes bonding sites are interchangeable by free rotation about the C ¬ C single bonds. In alkynes there is only one additional bonding site on a triply bound carbon, so no isomerism results. 24.33 (a) No H (b)
H
H
CH
(d) 2,2,5-trimethylhexane (e) methylcyclobutane 24.19 65 24.21 (a) Alkanes are said to be saturated because they cannot undergo addition reactions, such as those characteristic of carbon–carbon double bonds. (b) No. The compound C4H6 does not contain the maximum possible number of hydrogen atoms and is unsaturated. 24.23 (a) C5H12 (b) C5H10 (c) C5H10 (d) C5H8; saturated: (a), (b); unsaturated: (c), (d) 24.25 One possible structure is CH C CH CH C CH 24.27 There are at least 46 structural isomers with the formula C6H10. A few of them are
(d) CH3CH2CH2CH2CH(CH2CH3)CH(CH3)CH(CH3)2 H
CH2
CH2CH3
(b) sodium dicarbonyltetracyanoferrate(II) (c) +2, 6 d-electrons (d) We expect the complex to be low spin. Cyanide (and carbonyl) are high on the spectrochemical series, which means the complex will have a large ¢ splitting, characteristic of low-spin complexes. 23.91 (a) Yes, the oxidation state of Co is +3 in both complexes. (b) Compound A has SO42- outside the coordination sphere and coordinated Br- , so it forms a precipitate with BaCl 2(aq) but not AgNO3(aq). Compound B has Br- outside the coordination sphere and coordinated SO42 - , so it forms a precipitate with AgNO3(aq) but not BaCl 2(aq). (c) Compounds A and B are coordination sphere isomers. (d) Both compounds are strong electrolytes. 23.94 The chemical formula is 3Pd(NC5H5)2Br24. This is an electrically neutral square-planar complex of Pd(II), a nonelectrolyte whose solutions do not conduct electricity. Because the dipole moment is zero, it must be the trans isomer. 23.96 47.3 mg Mg2 + >L, 53.4 mg Ca2 + >L 23.99 ¢E = 3.02 * 10-19 J>photon, l = 657 nm. The complex will absorb in the visible around 660 nm and appear blue-green.
(e)
CH3 C
CN
Fe NC
(c) H3C
Cl C
ClH2C
ClH2C
C
Cl C
CH3
H
C CH3
(c) no (d) no 24.35 (a) An addition reaction is the addition of some reagent to the two atoms that form a multiple bond. In a substitution reaction one atom or group of atoms replaces another atom. Alkenes typically undergo addition, while aromatic hydrocarbons usually undergo substitution.
A-31
Answers to Selected Exercises (b)
24.49
CH3CH2CH CH CH3 Br2 2-pentene CH3CH2CH(Br)CH(Br)CH3 2, 3-dibromopentane
(a) CH3CH2O
O
H
C
(b) CH3N
O
Cl
H H
H
H
H
H
H FeCl3
C6H6 Cl2
H
O (a) CH3CH2C
C6H4Cl2
24.37 (a) The 60° C ¬ C ¬ C angles in the cyclopropane ring cause strain that provides a driving force for reactions that result in ring opening. There is no comparable strain in the five-or sixmembered rings. (b) C2H4(g) + HBr(g) ¡ CH3CH2Br(l); AlCl3
C6H6(l) + CH3CH2Br(l) ¡ C6H5CH2CH3(l) + HBr(g) 24.39 Not necessarily. That the two rate laws are first order in both reactants and second order overall indicates that the activated complex in the rate-determining step in each mechanism is bimolecular and contains one molecule of each reactant. This is usually an indication that the mechanisms are the same, but it does not rule out the possibility of different fast steps or a different order of elementary steps. 24.41 ¢Hcomb>mol CH2 for cyclopropane = 696.3 kJ, for cyclopentane = 663.4 kJ. ¢Hcomb>CH2 group for cyclopropane is greater because C3H6 contains a strained ring. When combustion occurs, the strain is relieved and the stored energy is released. 24.43 (a) Alcohol (b) amine, alkene (c) ether (d) ketone, alkene (e) aldehyde (f) carboxylic acid, alkyne 24.45 (a) Propionaldehyde (or propanal):
H
H
H
C
C
H
O
CH3CH2C
O
O (b) CH3C
NaOH
O
O CH3C
Na
O OH
24.53 The presence of both ¬ OH and ¬ C “ O groups in pure acetic acid leads us to conclude that it will be a strongly hydrogenbonded substance. That the melting and boiling points of pure acetic acid are both higher than those of water, a substance we know to be strongly hydrogen-bonded, supports this conclusion. 24.55 (a) CH3CH2CH2CH(OH)CH3 (b) CH3CH(OH)CH2OH (c) O CH3COCH2CH3
(d) H
H
C
C
H
H
H O
C
O H
C
H
(e) CH3OCH2CH3 24.57
O C
CH3 NaOH
Na CH3OH
H
H
O
O
C
(b) ethylmethyl ether:
H
CCH3
24.51
Cl
H
H
O
Phenylacetate
H
24.47 (a)
(c)
FeCl3
Cl2
OH
(b)
H
O CH3CH2CH2CH2C
(c)
CCH3
N-methylethanamide or N-methylacetamide
Ethylbenzoate
(c)
O
OH
O OH
CH3CH2CH2CH2CH2CH2CH2CH
C
H
CH3 Br
H
C
C
C*
C*
C
H
H
H
Cl
H
H *chiral C atoms
24.59 (a) An a -amino acid contains an NH2 group attached to the carbon adjacent to the carboxcylic acid function. (b) In protein formation, amino acids undergo a condensation reaction between the amino group of one molecule and the carboxylic acid group of another to form the amide linkage. (c) The bond that links amino acids in proteins is called the peptide bond. O
or
CH3 Cl
H
O C
C OH
N H
A-32
Answers to Selected Exercises
24.61
CH3
CH
CH2
CH2
CH3
NH3 O
H
C
N
C
O CH
C
O
H2C
H
N
N
CH2
NH3 O
H
H
O
C
N
C
C
C
O
CH2
H
CH CH3
24.73 Two important kinds of lipids are fats and fatty acids. Structurally, fatty acids are carboxylic acids with a hydrocarbon chain of more than four carbon atoms (typically 16–20 carbon atoms). Fats are esters formed by condensation of an alcohol, often glycerol, and a fatty acid. Phospholipids are glycerol esters formed from one phosphoric acid [RPO(OH)2] and two fatty acid (RCOOH) molecules. At body pH, the phosphate group is depronated and has a negative charge. The long, nonpolar hydrocarbon chains do not readily mix with water, but they do interact with the nonpolar chains of other phospholipid molecules to form the inside of a bilayer. The charged phosphate heads interact with polar water molecules on the outsides of the bilayer. 24.75 Purines, with the larger electron cloud and molar mass, will have larger dispersion forces than pyrimidines in aqueous solution. 24.77 5¿ -TACG -3¿ 24.79 The complimentary strand for 5¿ -GCATTGGC-3¿ is 3¿ -CGTAACCG-5¿ . 24.81 H
CH3
H 2C
24.63 (a) O
H
O
C
C
24.83 H
CH2 N
CH2CH3
CH3 cis
(b) Three tripeptides ar possible: Gly-Gly-His, GGH; Gly-His-Gly, GHG; His-Gly-Gly, HGG 24.65 The primary structure of a protein refers to the sequence of amino acids in the chain. The secondary structure is the configuration (helical, folded, open) of the protein chain. The tertiary structure is the overall shape of the protein determined by the way the segments fold together. (b) X-ray crystallography is the primary and preferred technique for determining protein structure. 24.67 (a) Carbohydrates, or sugars, are polyhydroxyaldehydes or ketones composed of carbon, hydrogen, and oxygen. They are derived primarily from plants and are a major food source for animals. (b) A monosaccharide is a simple sugar molecule that cannot be decomposed into smaller sugar molecules by hydrolysis. (c) A disaccharide is a carbohydrate composed of two simple sugar units. Hydrolysis breaks a disaccharide into two monosaccharides. (d) A polysaccharide is a polymer composed of many simple sugar units. 24.69 The empirical formula of cellulose is C6H10O5. As in glycogen, the six-membered ring form of glucose forms the monomer unit that is the basis of the polymer cellulose. In cellulose, glucose monomer units are joined by b linkages. 24.71 (a) In the linear form of mannose, the aldehydic carbon is C1. Carbon atoms 2, 3, 4, and 5 are chiral because they each carry four different groups. (b) Both the a (left) and b (right) forms are possible. CH2OH
6
CH2OH
H
C
O
5
C
O
C
H OH
OH C
H OH
1 OH C
H
H
OH
H
N
OH
C
C
C
OH
C
H
H3NCH2CNHCH2CNHCHCO
C
H H
O
O
OH C
C
H 4C
OH C 3 H
OH
C2 H H
CH3
H C
C CH2CH3
H trans
Cyclopentene does not show cis-trans isomerism because the existence of the ring demands that the C ¬ C bonds be cis to one another. 24.86 (a) Aldehyde, trans-alkene, cis-alkene (b) ether, alcohol, alkene, amine (two of these, one aliphatic and one aromatic) (c) ketone (two of these), amine (two of these) (d) amide, alcohol (aromatic) 24.88 In a carboxylic acid, the electronegative carbonyl oxygen withdraws electron density from the O ¬ H bond, rendering the bond more polar and the H more ionizable. And carboxylate anion is stabilized by resonance and encourages ionization of the carboxylic acid. In an alcohol no electronegative atoms are bound to the carbon that holds the ¬ OH group, and the H is tightly bound to the O. 24.92 Glu-CysGly is the only possible order. Glutamic acid has two carboxyl groups that can form a peptide bond with cysteine, so there are two possible structures for glutathione. 24.95 In both cases, stronger intermolecular forces lead to the higher boiling point. Ethanol contains O ¬ H bonds, which form strong intermolecular hydrogen bonds, while dimethyl ether experiences only weak dipole–dipole and dispersion forces. The heavier and polar CH2F2 experiences dipole–dipole and stronger dispersion forces, while CH4 experiences only weaker dispersion forces. 24.97 24.99 ¢G° = 13 kJ O CH3CCH2CH3
ANSWERS TO GIVE IT SOME THOUGHT CHAPTER 1 page 5 (a) 100 (b) atoms page 10 Water is composed of two types of atoms: hydrogen and oxygen. Hydrogen is composed only of hydrogen atoms, and oxygen is composed only of oxygen atoms. Therefore, hydrogen and oxygen are elements and water is a compound. page 13 (a) Chemical change: Carbon dioxide and water are different compounds than sugar. (b) Physical change: Water in the gas phase becomes water in the solid phase (frost). (c) Physical change: Gold in the solid state becomes liquid and then resolidifies. page 16 pg, picogram (10-12 g) page 19 2.5 * 102 m3 is, because it has units of length to the third power. page 21 (b) Mass of a penny page 27 Use all digits given in the conversion factor. Conversion factors may be exact and then have “infinite” significant digits (for example, 2.54 cm = 1 inch exactly). Usually, your answer will have its number of significant digits limited by those of the quantities given in the problem.
CHAPTER 2 page 41 (a) The law of multiple proportions. (b) The second compound must contain two oxygen atoms for each carbon atom (that is, twice as many carbon atoms as the first compound). page 44 Most a particles pass through the foil without being deflected because most of the volume of the atoms that comprise the foil is empty space. page 45 (a) The atom has 15 electrons because atoms have equal numbers of electrons and protons. (b) The protons reside in the nucleus of the atom. page 48 Any single atom of chromium must be one of the isotopes of that element. The isotope mentioned has a mass of 52.94 amu and is probably 53Cr. The atomic weight differs from the mass of any particular atom because it is the average atomic mass of the naturally occurring isotopes of the element. page 51 (a) Cl, (b) third period and group 7A, (c) 17, (d) nonmetal page 54 (a) C2H6, (b) CH3, (c) Probably the ball-and-stick model because the angles between the sticks indicate the angles between the atoms page 57 We write the empirical formulas for ionic compounds. Thus, the formula is CaO. page 60 (a) The transition metals can form more than one type of cation, and the charges of these ions are therefore indicated explicitly with Roman numerals: Chromium(II) ion is Cr2 + . Calcium, on the other hand, always forms the Ca2 + ion, so there is no need to distinguish it from other calcium ions with different charges. (b) The –ium ending indicates that the ion is formed from nonmetals. page 61 An -ide ending usually means a monatomic anion, although there are some anions with two atoms that are also named this way. An -ate ending indicates an oxyanion. The most common oxyanions have the -ate ending. An -ite ending also indicates an oxyanion, but one having less O than the anion whose name ends in -ate. page 62 BO33 - and SiO44 - . The borate has three O atoms, like the other oxyanions of the second period in Figure 2.27, and its charge is 3 - , following the trend of increasing negative charge as you move to the left in the period. The silicate has four O atoms, as do the other oxyanions in the third period in Figure 2.25, and its charge is 4- , also following the trend of increasing charge moving to the left.
page 65 Iodic acid, by analogy to the relationship between the chlorate ion and chloric acid page 67
H
H
H
H
H
C
C
C
C
H
H
H
H
H
H
H
H
H
C
C
C
H H
C
H H
H
H
CHAPTER 3 page 78 Each Mg(OH)2 has 1 Mg, 2 O, and 2 H; thus, 3 Mg(OH)2 represents 3 Mg, 6 O, and 6 H. page 83 The product is an ionic compound involving Na+ and S2-, and its chemical formula is therefore Na2S. page 88 (a) A mole of glucose. By inspecting their chemical formulas we find that glucose has more atoms of H and O than water and in addition it also has C atoms. Thus, a molecule of glucose has a greater mass than a molecule of water. (b) They both contain the same number of molecules because a mole of each substance contains 6.02 * 1023 molecules. page 93 The N:H ratio is 2:4 = 1:2. page 96 There are experimental uncertainties in the measurements. page 97 3.14 mol because 2 mol H2 ] 1 mol O2 based on the coefficients in the balanced equation page 98 The number of grams of product formed is the sum of the masses of the two reactants, 50 g. When two substances react in a combination reaction, only one substance is formed as a product. According to the law of conservation of mass, the mass of the product must equal the masses of the two reactants.
CHAPTER 4 page 118 (a) K + (aq) and CN-(aq), (b) Na + (aq) and ClO4-(aq) page 119 NaOH because it is the only solute that is a strong electrolyte page 123 Na + (aq) and NO3-(aq) page 125 Three. Each COOH group will partially ionize in water to form H + (aq). page 126 Only soluble metal hydroxides are classified as strong bases and Al(OH)3 is insoluble. page 130 SO2(g) page 133 (a) -3, (b) +5 page 136 (a) Yes, nickel is below zinc in the activity series so Ni2 + (aq) will oxidize Zn(s) to form Ni(s) and Zn2 + (aq). (b) No reaction will occur because the Zn2 + (aq) ions cannot be further oxidized. page 139 The second solution is more concentrated, 2.50 M, than the first solution, which has a concentration of 1.00 M. page 142 The concentration is halved to 0.25 M.
CHAPTER 5 page 162 No. The potential energy is lower at the bottom of the hill. (b) Once the bike comes to a stop, its kinetic energy is zero, just as it was at the top of the hill. page 163 Open system. Humans exchange matter and energy with their surroundings. page 167 Endothermic
A-33
A-34
Answers to Give It Some Thought
page 169 The balance (current state) does not depend on the ways the money may have been transferred into the account or on the particular expenditures made in withdrawing money from the account. It depends only on the net total of all the transactions. page 169 Because E, P, and V are state functions that don’t depend on path, H = E + PV must also be a state function. page 170 No. If ¢V is zero, then the expression w = - P¢V is also zero. page 171 A thermometer to measure temperature changes page 173 No. Because only half as much matter is involved, the value of ¢H would be 12(-483.6 kJ) = -241.8 kJ. q page 176 Hg(l). Rearranging Equation 5.22 gives ¢T = . Cs * m When q and m are constant for a series of substances, then constant . Therefore, the element with the smallest Cs in ¢T = Cs Table 5.2 has the largest ¢T, Hg(l). page 181 (a) The sign of ¢H changes. (b) The magnitude of ¢H doubles. page 184 No. Because O3(g) is not the most stable form of oxygen at 25 °C, 1 atm 3O2(g) is4, ¢Hf° for O3(g) is not necessarily zero. In Appendix C we see that it is 142.3 kJ> mol. page 189 Fats, because they have the largest fuel value of the three page 191 Combustion of H2(g) produces only H2O(g). No CO2(g) or other gases that might contribute to climate change issues are produced.
CHAPTER 6 page 210 No. Both visible light and X-rays are forms of electromagnetic radiation. They therefore both travel at the speed of light, c. Their differing ability to penetrate skin is due to their different energies, which we will discuss in the next section. page 211 E = hn = (6.63 * 10-34 J-s)(5 * 10-3 s -1) = 3 * 10-30 J; this radiation cannot produce a burst of 5 * 10-36 J because it can only produce energy in multiples of 3 * 10-30 J. page 212 Ultraviolet. Figure 6.4 shows that a photon in the ultraviolet region of the electromagnetic spectrum has a higher frequency and therefore a greater energy than a photon in the infrared region. page 214 According to the third postulate, photons of only certain allowed frequencies can be absorbed or emitted as the electron changes energy state. The lines in the spectrum correspond to the allowed frequencies. page 215 Absorb, because it is moving from a lower-energy state (n = 3) to a higher-energy state (n = 7) page 217 Yes, all moving objects produce matter waves, but the wavelengths associated with macroscopic objects, such as the baseball, are too small to allow for any way of observing them. page 219 The small size and mass of subatomic particles. The term h> 4p in the uncertainty principle is a very small number that becomes important only when considering extremely small objects, such as electrons. page 220 Bohr proposed that the electron in the hydrogen atom moves in a well-defined circular path around the nucleus (an orbit). In the quantum-mechanical model, no effort is made to describe the motion of the electron. An orbital is a wave function related to the probability of finding the electron at any point in space. page 221 The energy of an electron in the hydrogen atom is proportional to -1>n2, as seen in Equation 6.5. The difference between -1>(2)2 and -1>(1)2 is much greater than the difference between -1>(3)2 and -1>(2)2. page 226 (a) There is one 3s orbital, three 3p orbitals, and ten 3d orbitals, for a total of 14 orbitals. (b) 3s 6 3p 6 3d. page 232 The 6s orbital, which starts to hold electrons at element 55, Cs
page 237 We can’t conclude anything! Each of the three elements has a different valence electron configuration for its (n - 1)d and ns subshells: For Ni, 3d84s2; for Pd, 4d10; and for Pt, 5d96s1.
CHAPTER 7 page 251 Atomic number is governed by the number of protons in the nucleus, but atomic weight is governed by both the number of protons and neutrons in the nucleus (electrons are too light to worry about). Co> Ni, Cu> Zn, and Te> I are other pairs of elements whose atomic weights are “off ” compared to their atomic numbers. page 254 The 2p electron in a Ne atom would experience a larger Zeff than the 3s electron in Na, due to the better shielding by all the 2s and 2p electrons for Na’s 3s electron. page 256 These trends work against each other: Zeff increasing would imply that the valence electrons are pulled tighter in to make the atom smaller, while orbital size “increasing” would imply that atomic size would also increase. The orbital size effect is larger: As you go down a column in the periodic table, atomic size generally increases. page 259 It is harder to remove another electron from Na + , so the process in Equation 7.3 would require more energy and, hence, shorter-wavelength light (see Sections 6.1 and 6.2). page 260 Since Zeff increases as you go from boron to carbon, we would expect that the first ionization energy would be larger for carbon. Therefore, I2 for C is even greater. page 262 The same page 264 The numbers are the same; the signs are opposite. page 265 Increasing metallic character is correlated with decreasing ionization energy. page 268 Since the melting point is so low, we would expect a molecular rather than ionic compound. Thus, so PCl3 is more likely than ScCl3. page 270 Its low ionization energy page 272 In the acidic environment of the stomach, carbonate can react to give carbonic acid, which decomposes to water and carbon dioxide gas. page 274 The longest wavelength of visible light is about 750 nm (Section 6.1). We can assume that this corresponds to the lowest energy of light (since E = hc>l ) needed to break bonds in hydrogen peroxide. If we plug in 750 nm for l , we can calculate the energy to break one OO bond in one molecule of hydrogen peroxide, in joules. If we multiple by Avogadro’s number, we can calculate how many joules it would take to break a mole of OO bonds in hydrogen peroxide (which is the number one normally finds). page 275 The halogens all have ground-state electron configurations that are ns2np5; sharing an electron with only one other atom makes stable compounds. page 276 We can estimate the radius to be 1.5 Å, and the first ionization energy to be 900 kJ> mol. In fact, its bonding radius is indeed 1.5 Å, and the experimental ionization energy is 920 kJ> mol.
CHAPTER 8 page 290 No. Cl has seven valence electrons. The first and second Lewis symbols are both correct—they both show seven valence electrons, and it doesn’t matter which of the four sides has the single electron. The third symbol shows only five electrons and is incorrect. page 292 CaF2 is an ionic compound consisting of Ca2 + and F- ions. When Ca and F2 react to form CaF2, each Ca atom loses two electrons to form a Ca2 + ion and each fluorine atom in F2 takes up an electron, forming two F- ions. Thus, we can say that each Ca atom transfers one electron to each of two fluorine atoms.
Answers to Give It Some Thought page 292 No. Figure 7.9 shows that the alkali metal with the smallest first ionization energy is Cs with a value of + 376 kJ>mol. Figure 7.11 shows that the halogen with the largest electron affinity is Cl with a value of -349 kJ>mol. The sum of the two energies gives a positive energy (endothermic). Therefore, all other combinations of alkali metals with halogens will also have positive values. page 296 Rhodium, Rh page 297 Weaker. In both H2 and H2 + the two H atoms are principally held together by the electrostatic attractions between the nuclei and the electron(s) concentrated between them. H2 + has only one electron between the nuclei whereas H2 has two and this results in the H ¬ H bond in H2 being stronger. page 298 Triple bond. CO2 has two C ¬ O double bonds. Because the C ¬ O bond in carbon monoxide is shorter, it is likely to be a triple bond. page 299 Electron affinity measures the energy released when an isolated atom gains an electron to form a 1- ion. The electronegativity measures the ability of the atom to hold on to its own electrons and attract electrons from other atoms in compounds. page 300 Polar covalent. The difference in electronegativity between S and O is 3.5 - 2.5 = 1.0. Based on the examples of F2, HF, and LiF, the difference in electronegativity is great enough to introduce some polarity to the bond but not sufficient to cause a complete electron transfer from one atom to the other. page 302 IF. Because the difference in electronegativity between I and F is greater than that between Cl and F, the magnitude of Q should be greater for IF. In addition, because I has a larger atomic radius than Cl, the bond length in IF is longer than that in ClF. Thus, both Q and r are larger for IF and, therefore, m = Qr will be larger for IF. page 303 Smaller dipole moment for C ¬ H. The magnitude of Q should be similar for C ¬ H and H ¬ I bonds because the difference in electronegativity for each bond is 0.4. The C ¬ H bond length is 1.1 Å and the H ¬ I bond length is 1.6 Å. Therefore m = Qr will be greater for H ¬ I because it has a longer bond (larger r). page 304 OsO4. The data suggest that the yellow substance is a molecular species with its low melting and boiling points. Os in OsO4 has an oxidation number of +8 and Cr in Cr2O3 has an oxidation number of +3. In Section 8.4, we learn that a compound with a metal in a high oxidation state should show a high degree of covalence and OsO4 fits this situation. page 308 There is probably a better choice of Lewis structure than the one chosen. Because the formal charges must add up to 0 and the formal charge on the F atom is +1, there must be an atom that has a formal charge of -1. Because F is the most electronegative element, we don’t expect it to carry a positive formal charge. page 310 Yes. There are two resonance structures for ozone that each contribute equally to the overall description of the molecule. Each O ¬ O bond is therefore an average of a single bond and a double bond, which is a “one-and-a-half ” bond. page 310 As “one-and-a-third” bonds. There are three resonance structures, and each of the three N ¬ O bonds is single in two of those structures and double in the third. Each bond in the actual ion is an average of these: (1 + 1 + 2)>3 = 113 . page 312 No, it will not have multiple resonance structures. We can’t “move” the double bonds, as we did in benzene, because the positions of the hydrogen atoms dictate specific positions for the double bonds. We can’t write any other reasonable Lewis structures for the molecule. page 312 The formal charge of each atom is shown here: F.C.
N
O
N
O
0
0
⫺1
⫹1
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The first structure shows each atom with a zero formal charge and therefore it is the dominant Lewis structure. The second one shows a positive formal charge for an oxygen atom, which is a highly electronegative atom, and this is not a favorable situation. page 315 The atomization of ethane produces 2 C(g) + 6 H(g). In this process, six C ¬ H bonds and one C ¬ C bond are broken. We can use 6D(C ¬ H) to estimate the amount of enthalpy needed to break the six C ¬ H bonds. The difference between that number and the enthalpy of atomization is an estimate of the bond enthalpy of the C ¬ C bond, D(C ¬ C). page 315 H2O2. From Table 8.4, the bond enthalpy of the O ¬ O single bond in H2O2 (146 kJ> mol) is much lower than that of the O “ O bond in O2 (495 kJ> mol). The weaker bond in H2O2 is expected to make it more reactive than O2.
CHAPTER 9 page 334 Octahedral. Removing two atoms that are opposite each other leads to a square-planar geometry. page 335 The molecule does not follow the octet rule because it has ten electrons around the central A atom. There are four electron domains around A: two single bonds, one double bond, and one nonbonding pair. page 336 Each of the three represents a single electron domain in the VSEPR model. page 339 Yes. Based on one resonance structure, we might expect the electron domain that is due to the double bond to “push” the domains that are due to the single bonds, leading to angles slightly different from 120°. However, we must remember that there are two other equivalent resonance structures—each of the three O atoms has a double bond to N in one of the three resonance structures (Section 8.6). Because of resonance, all three O atoms are equivalent, and they will experience the same amount of repulsion, which leads to bond angles equal to 120°. page 339 A tetrahedral arrangement of electron domains is preferred because the bond angles are 109.5° compared to 90°. bond angles in a square-planar arrangement of electron domains. The larger bond angles result in smaller repulsions among electron domains and a more stable structure. page 343 Yes. The C ¬ O and C ¬ S bond dipoles exactly oppose each other, like in CO2, but because O and S have different electronegativities, the magnitudes of the bond dipoles will be different. As a consequence, the bond dipoles will not cancel each other and the OCS molecule has a nonzero dipole moment. page 348 Both p orbitals are perpendicular to the Be ¬ F bond axes. page 348 (bottom) The unhybridized p orbital is oriented perpendicular to the plane defined by the three sp2 hybrids (trigonal-planar array of lobes) with one lobe on each side of the plane. page 353 The molecule should not be linear. Because there are three electron domains around each N atom, we expect sp2 hybridization and H ¬ N ¬ N angles of approximately 120°. The molecule is expected to be planar; the unhybridized 2p orbitals on the N atoms can form a p bond only if all four atoms lie in the same plane. You might notice that there are two ways in which the H atoms can be arranged: They can be both on the same side of the N “ N bond or on opposite sides of the N “ N bond. page 358 The s bond component is formed from sp hybrid orbitals. page 360 The molecule would fall apart. With one electron in the bonding MO and one in the antibonding MO, there is no net stabilization of the electrons relative to two separate H atoms. page 362 Yes. In Be 2 + there would be two electrons in the s2s MO but only one electron in the s* 2s MO; therefore, the ion is predicted to have a bond order of 12 . It should (and does) exist. page 366 No. If the s2p MO were lower in energy than the p2p MOs, we would expect the s2p MO to hold two electrons and the p2p MOs to hold one electron each, with the same spin. The molecule would therefore be paramagnetic.
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Answers to Give It Some Thought
CHAPTER 10 page 384 Small page 385 1470 lb page 389 It would be halved. page 390 No—you have to convert T to Kelvin to calculate this properly. page 392 Avogadro’s number, 6.022 * 1023 page 396 Less dense page 399 The pressure due to N2 would be the same, but the total pressure would increase. page 404 HCl (slowest) 6 O2 6 H2 (fastest) page 406 3> 2 page 409 (a) Decrease, (b) No change page 410 (b) 100 K and 5 atm page 411 They do have intermolecular attractions for each other, and they do take up space.
CHAPTER 11 page 430 CH4 6 CCl4 6 CBr4. Because all three molecules are nonpolar, the strength of dispersion forces determines the relative boiling points. Polarizability increases in order of increasing molecular size and molecular weight, CH4 6 CCl4 6 CBr4; hence, the dispersion forces and boiling points increase in the same order. page 434 Ca(NO3)2 in water, because calcium nitrate is a strong electrolyte that forms ions and water is a polar molecule with a dipole moment. Ion–dipole forces cannot be present in a CH3OH> H2O mixture because CH3OH does not form ions. page 438 (a) Both viscosity and surface tension decrease with increasing temperature because of the increased molecular motion. (b) Both properties increase as the strength of intermolecular forces increases. page 440 Melting (or fusion), endothermic page 443 CCl4. Both compounds are nonpolar; therefore, only dispersion forces exist between the molecules. Because dispersion forces are stronger for the larger, heavier CBr4, it has a lower vapor pressure than CCl4. The substance with the larger vapor pressure at a given temperature is more volatile.
CHAPTER 12 page 466 Tetragonal. There are two three-dimensional lattices that have a square base with a third vector perpendicular to the base, tetragonal and cubic, but in a cubic lattice the a, b, and c lattice vectors are all of the same length. page 473 The packing efficiency decreases as the number of nearest neighbors decreases. The structures with the highest packing efficiency, hexagonal and cubic close packing, both have atoms with a coordination number of 12. Body-centered cubic packing, where the coordination number is 8, has a lower packing efficiency, and primitive cubic packing, where the coordination number is 6, has a lower packing efficiency still. page 474 Interstitial, because boron is a small nonmetal atom that can fit in the voids between the larger palladium atoms page 481 (a) Gold, Au. Tungsten, W, lies near the middle of the transition metal series where the bands arising from the d orbitals and the s orbital are approximately half-filled. This electron count should fill the bonding orbitals and leave the antibonding orbitals mostly empty. (b) Because both elements have similar numbers of electrons in the bonding orbitals but tungsten has fewer electrons in antibonding orbitals, it will have a higher melting point. page 482 No. In a crystal the lattice points must be identical. Therefore, if an atom lies on top of a lattice point, then the same type of atom
must lie on all lattice points. In an ionic compound there are at least two different types of atoms, and only one can lie on the lattice points. page 484 Four. The empirical formula of potassium oxide is K2O. Rearranging Equation 12.1 we can determine the potassium coordination number to be anion coordination number * (number of anions per formula unit> number of cations per formula unit) = 8(1>2) = 4. page 494 A condensation polymer. The presence of both ¬COOH and ¬NH2 groups allow molecules to react with one another forming C¬N bonds and splitting out H2O. page 495 As the vinyl acetate content increases more side chain branching occurs which inhibits the formation of crystalline regions thereby lowering the melting point. page 498 No. The emitted photons have energies that are similar in energy to the band gap of the semiconductor. If the size of the crystals is reduced into the nanometer range, the band gap will increase. However, because 340-nm light falls in the UV region of the electromagnetic spectrum, increasing the energy of the band gap will only shift the light deeper into the UV.
CHAPTER 13 page 514 Energy (or enthalpy) and entropy page 515 The lattice energy of NaCl(s) must be overcome to separate Na + and Cl- ions and disperse them into a solvent. C6H14 is nonpolar. Interactions between ions and nonpolar molecules tend to be very weak. Thus, the energy required to separate the ions in NaCl is not recovered in the form of ion–C6H14 interactions. page 517 (a) Separating solvent molecules from each other requires energy and is therefore endothermic. (b) Forming the solute– solvent interactions is exothermic. page 519 The added solute provides a template for the solid to begin to crystallize from solution, and the excess dissolved solute comes out of solution leaving a saturated solution. page 522 The solubility in water would be considerably lower because there would no longer be hydrogen bonding with water, which promotes solubility. page 526 Dissolved gases become less soluble as temperature increases, and they come out of solution, forming bubbles below the boiling point of water. page 526 230 ppm (1 ppm is 1 part in 106); 2.30 * 105 ppb (1 ppb is 1 part in 109). page 528 For dilute aqueous solutions the molality will be nearly equal to the molarity. Molality is the number of moles of solute per kilogram of solvent, whereas molarity is the number moles of solute per liter of solution. Because the solution is dilute, the mass of solvent is essentially equal to the mass of the solution. Furthermore, a dilute aqueous solution will have a density of 1.0 kg> L. Thus, the number of liters of solution and the number of kilograms of solvent will be essentially equal. page 531 The lowering of the vapor pressure depends on the total solute concentration (Equation 13.11). One mole of NaCl (a strong electrolyte) provides 2 mol of particles (1 mol of Na + and 1 mol Cl), whereas one mole of (a nonelectrolyte) provides only 1 mol particles. page 534 Not necessarily; if the solute is a strong or weak electrolyte, it could have a lower molality and still cause an increase of 0.51 °C. The total molality of all the particles in the solution is 1 m. page 537 The 0.20-m solution is hypotonic with respect to the 0.5-m solution. (A hypotonic solution will have a lower concentration and hence a lower osmotic pressure.) page 539 They would have the same osmotic pressure because they have the same concentration of particles. (Both are strong electrolytes that are 0.20 M in total ions.) page 543 The smaller droplets carry negative charges because of the embedded stearate ions and thus repel one another.
Answers to Give It Some Thought
CHAPTER 14 page 559 The rate will increase. page 562 Average rate is for a large time interval; instantaneous rate is for an “instant” in time. Yes, they can have the same numeric value, especially if a plot of concentration versus time is linear. page 565 (top) Reaction rate is what we measure as a reaction proceeds—change in concentration in time for one or more of the components in the mixture. Reaction rate always has units of concentration per time, usually M> s. A rate constant is what we calculate from reaction rate data, and its magnitude is proportional to the reaction rate, but its units depend on the reaction order. The rate law of a reaction is an equation that relates reaction rate to the rate constant: Rate = k3A4m3B4n, for components A and B in the reaction. page 565 No. Rate is always change in concentration per time; rate constant has units that depend on the form of the rate law. page 566 (a) The reaction is second order in NO, first order in H2, and third order overall. (b) No. Doubling NO concentration will quadruple the rate, but doubling H2 concentration will merely double the rate. page 567 No reaction will take place. page 573 1.25 g page 575 The half-life will increase. page 578 No—transition states are by definition not stable. page 578 The collision may not have occurred with enough energy for reaction to occur, and> or the collision may not have occurred with the proper orientation of reactant molecules to favor product formation. page 581 Bimolecular page 585 Most reactions occur in elementary steps; the rate law is governed by the elementary steps, not by their sum (which is the overall balanced equation). page 587 The odds of three molecules colliding with each other properly to react is very low. page 590 By lowering the activation energy for the reaction or by increasing the frequency factor page 591 A homogeneous catalyst will be harder to separate from the reaction mixture than a heterogeneous one. page 593 People do say this, but we have to be careful. An enzymecatalyzed reaction will have a lower transition state energy than the uncatalyzed reaction, but the nature of the transition state is probably different than the uncatalyzed version.
CHAPTER 15 page 614 (a) The rates of the forward and reverse reactions. (b) Greater than 1 page 614 When the concentrations of reactants and products are no longer changing page 617 It does not depend on starting concentrations. page 617 Units of moles> L are used to calculate Kc ; units of partial pressure are used to calculate Kp. page 618 0.00140 page 621 It is cubed. page 623 Kp = PH2O page 625 Kc = 3NH4 + 43OH-4>3NH34 page 633 (a) It shifts to the right. (b) It shifts to the left. page 633 (bottom) It will shift to the left, the side with a larger number of moles of gas. page 636 As the temperature increases, a larger fraction of molecules in the liquid phase have enough energy to overcome their inter-
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molecular attractions and go into the vapor; the evaporation process is endothermic. page 638 No
CHAPTER 16 page 652 The H + ion for acids and the OH- ion for bases page 654 CH3NH2 is the base because it accepts a H + from H2S as the reaction moves from the left-hand to the right-hand side of the equation. page 657 As the conjugate base of a strong acid, we would classify ClO4 - as having negligible basicity. page 661 pH is defined as -log3H + 4. This quantity will become negative if the H + concentration exceeds 1 M, which is possible. Such a solution would be highly acidic. page 662 pH = 14.00 - 3.00 = 11.00. This solution is basic because pH 7 7.0. page 665 Both NaOH and Ba(OH)2 are soluble hydroxides. Therefore, the hydroxide concentrations will be 0.001 M for NaOH and 0.002 M for Ba(OH)2. Because the Ba(OH)2 solution has a higher 3OH-4, it is more basic and has a higher pH. page 666 Because CH3 - is the conjugate base of a substance that has negligible acidity, CH3 - must be a strong base. Bases stronger than OH- abstract H + from water molecules: CH3 - + H2O ¡ CH4 + OH- . page 668 Oxygen page 671 Because weak acids typically undergo very little ionization, often less than 1%. Normally we make this assumption and then check its validity based on the concentration of conjugate base formed in the calculation. If it is …5% of the initial concentration of the weak acid, we can generally use this assumption. If not, we must do an exact calculation. page 674 This is the acid-dissociation constant for the loss of the third and final proton from H3PO4, corresponding to the equilibrium HPO42 - Δ H + + PO43 - . page 680 The pKa value is -log Ka = -log(6.8 * 10-4) = 3.17. The pKb value is 14.00 - pKa = 14.00 - 3.17 = 10.83. page 682 Nitrate is the conjugate base of nitric acid, HNO3. The conjugate base of a strong acid does not act as a base, so NO3 - ions will not affect the pH. Carbonate is the conjugate base of hydrogen carbonate, HCO3 - , which is a weak acid. The conjugate base of a weak acid acts as a weak base, so CO32 - ions will increase the pH. page 686 The increasing acidity going down a group is due mainly to decreasing H ¬ X bond strength. The trend going across a period is due mainly to the increasing electronegativity of X, which weakens the H ¬ X bond. page 687 HBrO3. For an oxyacid, acidity increases as the electronegativity of the central ion increases, which would make HBrO2 more acidic than HIO2. Acidity also increases as the number of oxygens bound to the central atom increases, which would make HBrO3 more acidic than HBrO2. Combining these two relationships we can order these acids in terms of increasing acid-dissociation constant, HIO2 6 HBrO2 6 HBrO3. page 689 The carboxyl group, ¬ COOH page 690 It must have an unshared pair of electrons that can be shared with another atom.
CHAPTER 17 page 707 (top) The Cl- ion is the only spectator ion. The pH is determined by the equilibrium NH3(aq) + H2O(l) ÷ OH-(aq) + NH4 + (aq). page 707 (bottom) HNO3 and NO3 -. To form a buffer we need comparable concentrations of a weak acid and its conjugate base.
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Answers to Give It Some Thought
HNO3 and NO3 - will not form a buffer because HNO3 is a strong acid and the NO3 - ion is merely a spectator ion. page 708 (a) The OH- of NaOH (a strong base) reacts with the acid member of the buffer (CH3COOH), abstracting a proton. Thus, 3CH3COOH4 decreases and 3CH3COO-4 increases. (b) The H + of HCl (a strong acid) reacts with the base member of the buffer 3CH3COOH - 4. Thus, 3CH3COO-4 decreases and 3CH3COOH4 increases. page 711 A buffer will be most resistant to changes in pH when the concentrations of the weak acid and its conjugate base are equal. When the two are exactly equal the Henderson–Hasselbach equation tells us that the pH of the buffer will be equal to the pKa of the weak acid. The pKa values of nitrous acid and hypochlorous acid are 3.35 and 7.52, respectively. Thus, HClO would be more suitable for a pH = 7.0 buffer solution. To make a buffer we would also need a salt containing ClO-, such as NaClO. page 716 The pH = 7. The neutralization of a strong base with a strong acid gives a salt solution at the equivalence point. The salt contains ions that do not change the pH of water. page 721 The following titration curve shows the titration of 25 mL of Na2CO3 with HCl, both with 0.1 M concentrations. The overall reaction between the two is Na2CO3(aq) + HCl(aq) ¡ 2 NaCl(aq) + CO2(g) + H2O(l) The initial pH (sodium carbonate in water only) is near 11 because CO32 - is a weak base in water. The graph shows two equivalence points, A and B. The first point, A, is reached at a pH of about 9: Na2CO3(aq) + HCl(aq) ¡ NaCl(aq) + NaHCO3(aq) HCO3 - is weakly basic in water and is a weaker base than the carbonate ion. The second point, B, is reached at a pH of about 4: NaHCO3(aq) + HCl(aq) ¡ NaCl(aq) + CO2(g) + H2O(l) H2CO3, a weak acid, forms and decomposes to carbon dioxide and water. 14
pH
A 7
B
0
0
25
50 Volume HCl (mL)
page 722 The nearly vertical portion of the titration curve at the equivalence point is smaller for a weak acid–strong base titration; as a result fewer indicators undergo their color change within this narrow range. page 724 AgCl. Because all three compounds produce the same number of ions, their relative stabilities correspond directly to the Ksp values, with the compound with the largest Ksp value being the most soluble. page 734 Amphoteric substances are insoluble in water but dissolve in the presence of sufficient acid or base. Amphiprotic substances can both donate and accept protons. page 738 The solution must contain one or more of the cations in group 1 of the qualitative analysis scheme, Ag + , Pb2 + or Hg 22 + .
CHAPTER 18 page 754 Photoionization is a process in which a molecule breaks into ions upon illumination with light; photodissociation is a process in which molecules break up upon illumination with light but the products bear no charge. page 755 Because those molecules do not absorb light at those wavelengths page 757 Yes—Cl is neither a product nor a reactant in the overall reaction, and its presence does speed the reaction up. page 760 SO2 in the atmosphere reacts with oxygen to form SO3. SO3 in the atmosphere reacts with water in the atmosphere to form H2SO4, sulfuric acid. The sulfuric acid dissolves in water droplets that fall to Earth, causing “acid rain” that has a pH of 4 or so. page 761 NO2 photodissociates to NO and O; the O atoms react with O2 in the atmosphere to form ozone, which is a key ingredient in photochemical smog. page 763 Higher humidity means there is more water in the air. Water absorbs infrared light, which we feel as heat. After sundown, the ground that has been warmed earlier in the day reradiates heat out. In locations with higher humidity, this energy is absorbed somewhat by the water and in turn is reradiated to some extent back to the Earth, resulting in warmer temperatures compared to a lowhumidity location. page 764 We need to be below water’s critical point. Therefore, to sublime water we need to be below 0.006 atm. A wide range of temperatures will work for sublimation at this pressure—the most environmentally relevant ones are - 50 °C to 100 °C. page 768 The pollutants are capable of being oxidized (either directly by reaction with dissolved oxygen or indirectly by the action of organisms such as bacteria). page 772 With a catalyst, the reaction is always faster, therefore costing less energy to run. In addition, with a catalyst the reaction may occur readily at a lower temperature, also costing less energy. page 773 Fossil fuel combustion puts a great deal more CO2 in the atmosphere right now than any supercritical use of CO2. Compared to other (halogenated organic) solvents, supercritical CO2 is far less toxic to life. Therefore, at present, using CO2 as a solvent or a reactant in industrial processes is a reasonable choice for environmental sustainability. page 774 Use room temperature and room pressure; use water as a solvent if possible; use O2 as the oxidizing agent instead of hydrogen peroxide if possible. page 775 sp before reaction; sp2 after reaction
CHAPTER 19 page 787 No, nonspontaneous processes can occur so long as they receive some continuous outside assistance. Examples of nonspontaneous processes with which we may be familiar include the building of a brick wall and the electrolysis of water to form hydrogen gas and oxygen gas. page 789 No. Just because the system is restored to its original condition doesn’t mean that the surroundings have likewise been restored to their original condition, so it is not necessarily reversible. page 791 ¢S depends not merely on q but on qrev. Although there are many possible paths that could take a system from its initial to final state, there is always only one reversible isothermal path between two states. Thus, ¢S has only one particular value regardless of the path taken between states. page 793 Because rusting is a spontaneous process, ¢Suniv must be positive. Therefore, the entropy of the surroundings must increase, and that increase must be larger than the entropy decrease of the system. page 795 S = 0, based on Equation 19.5 and the fact that ln 1 = 0.
Answers to Give It Some Thought page 796 A molecule can vibrate (atoms moving relative to one another) and rotate (tumble), whereas a single atom cannot undergo these motions. page 799 It must be a perfect crystal at 0 K (third law of thermodynamics), which means it has only a single accessible microstate. page 803 ¢Ssurr always increases. For simplicity, assume that the process is isothermal. The change in entropy of the surroundings in -qsys an isothermal process is ¢Ssurr = . Because the reaction is T exothermic, -qsys is a positive number. Thus, ¢Ssurr is a positive number and the entropy of the surroundings increases. page 805 (a) In any spontaneous process the entropy of the universe increases. (b) In any spontaneous process operating at constant temperature, the free energy of the system decreases. page 806 It indicates that the process to which the thermodynamic quantity refers has taken place under standard conditions, as summarized in Table 19.2. page 810 Above the boiling point, vaporization is spontaneous, and ¢G 6 0. Therefore, ¢H - T¢S 6 0, and ¢H 6 T¢S.
CHAPTER 20 page 829 Oxygen is first assigned an oxidation number of - 2. Nitrogen must then have a + 3 oxidation number for the sum of oxidation numbers to equal - 1, the charge of the ion. page 832 No. Electrons should appear in the two half-reactions but cancel when the half-reactions are added properly. page 839 Yes. A redox reaction with a positive standard cell potential is spontaneous under standard conditions. page 840 1 atm pressure of Cl2(g) and 1 M concentration of Cl-(aq) ° = -0.126 V for page 846 Using data from Appendix E, we have E red ° = 0.854 V for Hg2 + (aq) : Hg(l). Pb2 + (aq) : Pb(s) and E red ° , it is the stronger Because Pb(s) has the most negative value for E red reducing agent. (See Figure 20.12.) The comparison can also be made by reference to the activity series where Pb lies also above Hg, indicating that Pb is oxidized more readily than Hg. The more readily a substance is oxidized, the stronger it is as a reducing agent. page 859 Al, Zn. Both are easier to oxidize than Fe.
CHAPTER 21 page 877 The mass number decreases by 4. page 879 Only the neutron, as it is the only neutral particle listed. page 883 From Figure 21.4 we can see that each of these four elements has only one stable isotope, and from their atomic numbers we see that they each have an odd number of protons. Given the rarity of stable isotopes with odd numbers of neutrons and protons, we expect that each isotope will possess an even number of neutrons. From their atomic weights we see that this is the case: F (10 neutrons), Na (12 neutrons), Al (14 neutrons), and P (16 neutrons). page 885 No. Electric and magnetic fields are only effective at accelerating charged particles and a neutron is not charged. page 889 top Spontaneous radioactive decay is a unimolecular process: A : Products. The rate law that fits this observation is a first-order kinetic rate law, rate = k3A4. A second-order kinetic process has rate = k3A42 and the elementary reaction is bimolecular: A + A : Products. A zero-order kinetic process has rate = k, and the rate does not change until the limiting reactant is entirely consumed. The latter two rate laws do not fit a unimolecular process. page 889 (bottom) (a) Yes; doubling the mass would double the amount of radioactivity of the sample as shown in Equation 21.18. (b) No; changing the mass would not change the half-life as shown in Equation 21.20.
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page 892 No. Alpha particles are more readily absorbed by matter than beta or gamma rays. Geiger counters must be calibrated for the radiation they are being used to detect. page 896 (top) The values in Table 21.7 only reflect the mass of the nucleus, while the atomic mass is the sum of the mass of the nucleus and the electrons. So the atomic mass of iron-56 is 26 * me larger than the nuclear mass. page 896 (bottom) No. Stable nuclei having mass numbers around 100 are the most stable nuclei. They could not form a still more stable nucleus with an accompanying release of energy. page 905 The absorbed dose is equal to 0.10 J * (1 rad>1 * 10-2 J) = 10 rads. The effective dosage is calculated by multiplying the absorbed dose by the relative biological effectiveness (RBE) factor, which is 10 for alpha radiation. Thus, the effective dosage is 100 rems.
CHAPTER 22 page 919 (top) No. page 919 No. N can form triple bonds but P cannot, as it would have to form P2 . page 921 H-, hydride. page 923 +1 for everything except H2, for which the oxidation state of H is 0. page 924 No—it is the volume of Pd that can increase to accommodate hydrogen, not its mass. page 927 0 for Cl2; -1 for Cl- ; +1 for ClOpage 929 They should both be strong, since the central halogen is in the +5 oxidation state for both of them. We need to look up the redox potentials to see which ion, BrO3 - or ClO3 -, has the larger reduction potential. The ion with the larger reduction potential is the stronger oxidizing agent. BrO3 - is the stronger oxidizing agent on this basis ( +1.52 V standard reduction potential in acid compared to +1.47 V for ClO3 - ). page 931 The standard energy to dissociate one mole of oxygen atoms from one mole of ozone was given as 105 kJ. If we assume, as usual, that one photon will dissociate one molecule, that means the energy of the photons should be 105 kJ per mole (of photons). Using Avogadro’s number, we can calculate that one photon would then have 1.744 * 10-19 J of energy. Using equations from Chapter 6, c = ln and E = hn, we can find that a photon with 1.744 * 10-19 J of energy will have a wavelength l of 1140 nm, or 1.14 * 106 m, which is in the infrared part of the spectrum. page 932 HIO3 page 936 SO3(g) + H2O(l) ¡ H2SO4(l) page 940 (a) +5 (b) +3 page 948 CO2(g) page 949 Yes, it must, since CS2 is a liquid at room temperature and pressure, and CO2 is a gas. page 952 Silicon is the element, Si. Silica is SiO2. Silicones are polymers that have an O ¬ Si ¬ O backbone and hydrocarbon groups on the Si. page 953 +3
CHAPTER 23 page 965 Sc is the biggest. page 967 You would have to remove core electrons. page 968 (top) The larger the distance, the weaker the spin–spin interactions. page 968 (bottom) Yes, it is a Lewis acid–base interaction; the metal ion is the Lewis acid (electron pair acceptor). page 972 3Fe(H2O)643 + (aq) + SCN-(aq) ¡ 3Fe(H2O)5SCN42 + (aq) + H2O(l)
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Answers to Give It Some Thought
page 974 (a) tetrahedral (b) octahedral page 976 Bidentate page 978 Its conjugation (alternating single and double CC bonds) page 981 No, ammonia cannot engage in linkage isomerism—the only atom that can coordinate to a metal is the nitrogen. page 985 Both isomers have the same chemical formulas and the same donor atoms on the ligands bonding to the metal ion. The difference is that the d isomer has a right-handed “twist” and the l isomer has a “left-handed” twist. page 987 Co is 1s22s22p63s23p64s23d7. Co3 + is 1s22s22p63s23p63d6. Co has 3 unpaired electrons; Co3 + has 4 unpaired electrons, assuming all 5 d orbitals have the same energy. page 989 It has lost all of the Ti valence electrons; only core electrons remain, and the energy gap between filled and empty orbitals is large, corresponding to light in the ultraviolet, which we cannot perceive as colored. page 991 Low spin page 992 The ligands are in the xy plane. The dxy orbital has its lobes mostly in that plane, so its energy is higher than dxz and dyz.
CHAPTER 24 page 1007 C “ N, because it is a polar double bond. C ¬ H and C ¬ C bonds are relatively unreactive. page 1009 Two C ¬ H bonds and two C ¬ C bonds
page 1010 The isomers have different properties, as seen in Table 24.3. page 1015 Only two of the four possible C “ C bond sites are distinctly different in the linear chain of five carbon atoms with one double bond. page 1021
NO2 NO2
page 1025
O H2C H2C
C CH2 CH2
page 1029 All four groups must be different from one another. page 1033 No. Breaking the hydrogen bonds between N ¬ H and O “ C groups in a protein by heating causes the a-helix structure to unwind and the b-sheet structure to separate. page 1037 The a form of the C ¬ O ¬ C linkage. Glycogen serves as a source of energy in the body, which means that the body’s enzymes must be able to hydrolyze it to sugars. The enzymes work only on polysaccharides having the a linkage.
ANSWERS TO GO FIGURE Figure 1.1 9 Figure 1.4 Vapor (gas) Figure 1.5 Molecules of a compound are composed of more than one type of atom, and molecules of an element are composed of only one type of atom. Figure 1.6 Earth is rich in silicon and aluminum; the human body is rich in carbon and hydrogen Figure 1.7 They are the same; there are twice as many hydrogen molecules as oxygen molecules, and the hydrogen gas takes up twice the volume of the oxygen gas. Figure 1.17 True Figure 1.18 1000 Figure 1.23 The darts would be scattered widely (poor precision) but their average position would be at the center (good accuracy).
CHAPTER 2 Figure 2.3 We know the rays travel from the cathode because of the way the magnetic field diverts the path (b). Figure 2.4 The electron beam would be deflected downward because of repulsion by the negative plate and attraction toward the positive plate. Figure 2.8 The beta rays, whose path is diverted away from the negative plate and toward the positive plate, consist of electrons. Because the electrons are much less massive than the alpha particles, their motion is affected more strongly by the electric field. Figure 2.10 The beam consists of alpha particles, which carry a +2 charge. Figure 2.14 Based on the periodic trend, we expect that elements that precede a nonreactive gas, as F does, will also be reactive nonmetals. The elements fitting this pattern are H and Cl. Figure 2.19 The ball-and-stick model more clearly shows the connections between atoms, so we can see the angles at which the atoms are attached in the molecule. Figure 2.20 The elements are in the following groups: Ag⫹ is 1B, Zn2⫹ is 2B, and Sc3⫹ is 3B. Sc3⫹ has the same number of electrons as Ar (element 18). Figure 2.24 Removing one O atom from the perbromate ion gives the bromate ion, BrO3⫺.
CHAPTER 3 Figure 3.3 The formula CO2 represents one molecule containing one C and two O atoms, whereas 2 CO represents two molecules, each containing one C atom and one O atom for a total of two C and two O atoms. Figure 3.8 Both figures show combustion reactions in which the fuel is a hydrocarbon (CH4 in Figure 3.4 and C3H8 in Figure 3.8). In both cases the reactants are the hydrocarbon and O2, and the products are CO2 and H2O. Figure 3.9 As shown, 18.0 g H2O = 1 mol H2O = 6.02 * 1023 molecules H2O. Thus, 9.00 g H2O = 0.500 mol H2O = 3 .01 * 1023 molecules H2O. Figure 3.12 (a) The molar mass of CH4, 16.0 g CH4/1 mol CH4. (b) Avogadro’s number, 1 mol CH4>6 .02 * 1023 formula units CH4, where a formula unit in this case is a molecule. Figure 3.13 The mole ratio is obtained by dividing the molecular weight by the empirical formula weight, Equation 3.11.
Figure 3.17 There are 7 mol O2, and each mol O2 yields 2 mol H2O. Thus, 14 mol H2O would have formed.
CHAPTER 4 Figure 4.3 NaCl(aq) Figure 4.4 K⫹ and NO3⫺ Figure 4.9 Two moles of hydrochloric acid are needed to react with each mole of Mg(OH)2. Figure 4.19 The volume needed to reach the end point if Ba(OH)2(aq) were used would be one-half the volume needed for titration with NaOH(aq).
CHAPTER 5 Figure 5.1 In the act of throwing, the pitcher transfers energy to the ball, which then becomes kinetic energy of the ball. For a given amount of energy E transferred to the ball, Equation 5.1 tells us that the speed of the ball is v = 12E>m where m is the mass of the ball. Because a baseball has less mass than a bowling bowl, it will have a higher speed for a given amount of energy transferred. Figure 5.2 When she starts going uphill, kinetic energy is converted to potential energy and her speed decreases. Figure 5.3 The electrostatic potential energy of two oppositely charged particles is negative (Equation 5.2). As the particles become closer, the electrostatic potential energy becomes even more negative— that is, it decreases. Figure 5.4 Yes, the system is still closed—matter can’t escape the system to the surroundings unless the piston is pulled completely out of the cylinder. Figure 5.5 If Efinal = Einitial, then ¢E = 0. Figure 5.6 Mg(s) ⫹ Cl2(g) Final state Internal energy, E
CHAPTER 1
⌬E ⬎ 0
Initial state
MgCl2(s)
Figure 5.7 No. The sign on w is positive and the sign on q is negative. We need to know the magnitudes of q and w to determine whether ¢E = q + w is positive or negative. Figure 5.10 The battery is doing work on the surroundings, so w 6 0. Figure 5.11 We need to know whether Zn(s) or HCl(aq) is the limiting reagent of the reaction. If it is Zn(s), then the addition of more Zn will lead to the generation of more H2(g) and more work will be done. Figure 5.17 Endothermic—heat is being added to the system to raise the temperature of the water. Figure 5.18 Two cups provide more thermal insulation so less heat will escape the system. Figure 5.19 The stirrer ensures that all of the water in the bomb is at the same temperature.
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Answers to Go Figure
Figure 5.21 The condensation of 2 H2O(g) to 2 H2O(l) Figure 5.22 Yes, ¢H3 would remain the same as it is the enthalpy change for the process CO(g) + 12 O2(g) ¡ CO2(g). Figure 5.24 Grams of fat
CHAPTER 6 Figure 6.3 The wavelength of (a) is twice that of (b) and the frequency of (a) is consequently half that of (b). Thus, the wavelength of (b) is 0.50 m and its frequency is 6 .0 * 108 cycles/s. Figure 6.4 The X-ray has a shorter wavelength and, consequently, higher frequency than the red light. Figure 6.5 The hottest area is the white or yellowish white area in the center. Figure 6.7 If the tube is not evacuated, the electrons that are freed from the metal surface will strike gas molecules near that surface. As a result, they will become attached to the gas molecules and never arrive at the positive terminal. Figure 6.12 The n = 2 to n = 1 transition involves a larger energy change than the n = 3 to n = 2 transition. (Compare the space differences between the states in the figure.) If the n = 2 to n = 1 transition produces visible light, the n = 3 to n = 2 transition must produce radiation of lower energy. The infrared radiation has lower frequency and, hence, lower energy than visible light, whereas the ultraviolet has greater frequency and greater energy. Thus, the n = 2 to n = 1 transition will produce ultraviolet radiation. Figure 6.16 The region of highest electron density is where the density of dots is highest, which is near the nucleus. Figure 6.17 The fourth shell (n = 4) would contain four subshells, labeled 4s, 4p, 4d, and 4f. Figure 6.18 There would be four maxima and three nodes. Figure 6.22 (a) The intensity of the color indicates that the probability of finding the electron is greater at the interior of the lobes than on the edges. (b) 2px. Figure 6.24 The 4d and 4f subshells are not shown.
CHAPTER 7 Figure 7.1 Row 7—these elements are generally radioactive and not stable. Figure 7.3 2s Figure 7.6 Bottom and left Figure 7.7 They get larger, just like the atoms do. Figure 7.9 Ar; it has a larger Zeff. Figure 7.10 There is more electron–electron repulsion in the case of oxygen because two electrons have to occupy the same orbital. Figure 7.11 The halogens (group 7A); it does make sense because we know that they are very stable as anions. Figure 7.12 Ionization energy—lower ionization energy is correlated with increasing metallic character. Figure 7.14 Anions are above the lines; cations are below the line. Figure 7.22 Lilac
CHAPTER 8 Figure 8.1 Covalent Figure 8.2 Yes, the same sort of reaction should occur between any of the alkali metals and any of the elemental halogens. Figure 8.3 Cations have a smaller radius than their neutral atoms and anions have a larger radius. Because Na and Cl are in the same row of the periodic table, we would expect Na⫹ to have a smaller radius than Cl⫺, so we would guess that the larger green spheres represent Cl⫺.
Figure 8.4 The distance between ions in KF should be larger than that in NaF and smaller than that in KCl. We would thus expect the lattice energy of KF to be between 701 and 910 kJ/mol. Figure 8.6 The repulsions between the nuclei would decrease, the attractions between the nuclei and the electrons would decrease, and the repulsions between the electrons would be unaffected. Figure 8.7 The electronegativity decreases with increasing atomic number. Figure 8.9 m will decrease Figure 8.10 The bonds are not polar enough to cause enough excess electron density on the halogen atom to lead to a red shading. Figure 8.12 The lengths of the bonds of the outer O atoms to the inner O atom are the same. Figure 8.13 Yes. The electron densities on the left and right parts of the molecule are the same, indicating that resonance has made the two O ¬ O bonds equivalent to one another. Figure 8.14 The dashed bonds represent the “half bonds” that result when the two resonance structures are averaged. Figure 8.15 Exothermic Figure 8.17 As the bond gets longer, it gets weaker. We would therefore expect a plot of bond enthalpy versus bond length to have a negative slope.
CHAPTER 9 Figure 9.1 The atomic radii (Figure 7.7) Figure 9.3 Octahedral Figure 9.7 The electron pair in the bonding domain is attracted toward two nuclear centers, whereas the nonbonding pair is attracted toward just one. Figure 9.8 90° Figure 9.9 The nonbonding electron pairs exert a greater repulsive force than the bonding electron pairs. Figure 9.10 The heads of the arrows point toward regions of highest electron density, as indicated by the red color. Figure 9.14 As the internuclear distance decreases, nucleus–nucleus repulsion becomes a dominant component of the potential energy. Figure 9.16 The small lobes of the sp hybrid orbitals are very much smaller in spatial extent and, therefore, provide very little overlap with the F orbitals. Figure 9.17 Three: one s and two p orbitals Figure 9.23 The two p orbitals that form the p bond must align, and each of them is perpendicular to the plane of the sp2 hybrid orbitals. Figure 9.24 Acetylene, because it has two C ¬ C p bonds, whereas ethylene has one p bond Figure 9.26 C ¬ H and C ¬ C Figure 9.33 The s*1s Figure 9.34 The two electrons in the s1s MO Figure 9.35 The 1s orbitals of Li are small in spatial extent because they experience a strong nuclear attraction. In addition, both the bonding and antibonding MOs formed from them are occupied, so that there is no significant net bonding. Figure 9.36 Nodal planes between the atoms are found in antibonding MOs. Figure 9.42 The s2p and p2p orbitals. Because the s2p orbital mixes with the s2s, it is pushed to higher energy and the s2s is moved to lower energy. The s2p orbital thus rises above the p2p in energy. Figure 9.43 F2 contains four more electrons than N2. These electrons go into the antibonding p*2p orbitals, thus lowering the bond order.
Answers to Go Figure Figure 9.45 Because N2 has no unpaired electrons, it is diamagnetic. Therefore, it would simply flow down with no tendency to remain in the magnetic field. Figure 9.46 11. All the electrons in the n = 2 level are valence-shell electrons.
CHAPTER 10 Figure 10.2 It will increase. Figure 10.5 Decrease Figure 10.6 1520 torr Figure 10.7 Linear Figure 10.10 one Figure 10.11 It is small and inert. Figure 10.17 About a third Figure 10.18 Higher speeds are correlated with smaller molar masses (assuming constant T). Figure 10.20 n, moles of gas Figure 10.22 Not really—CO2 is least ideal and does have the largest molar mass, but H2, the lightest gas, deviates more from the ideal line than the heavier N2. Figure 10.23 True Figure 10.25 It would increase.
CHAPTER 11 Figure 11.2 The density in a liquid is much closer to a solid than it is to a gas. Figure 11.9 Both compounds are nonpolar and incapable of forming hydrogen bonds. Therefore, the boiling point is determined by the dispersion forces, which are stronger for the larger, heavier SnH4. Figure 11.10 The non-hydrogen atom must possess a nonbonding electron pair. Figure 11.11 There are four electron pairs surrounding oxygen in a water molecule. Two of the electron pairs are used to make covalent bonds to hydrogen within the H2O molecule, while the other two are available to make hydrogen bonds to neighboring molecules. Because the electron-pair geometry is tetrahedral (four electron domains around the central atom), the H ¬ O Á H bond angle is approximately 109°. Figure 11.19 Wax is a hydrocarbon that cannot form hydrogen bonds. Therefore, coating the inside of tube with wax will dramatically decrease the adhesive forces between water and the tube and change the shape of the water meniscus to an inverted U-shape. Neither wax nor glass can form metallic bonds with mercury so the shape of the mercury meniscus will be qualitatively the same, an inverted U-shape. Figure 11.21 Because we are dealing with a state function, the energy of going straight from a solid to a gas must be the same as going from a solid to a gas through an intermediate liquid state. Therefore, the heat of sublimation must be equal to the sum of the heat of fusion and the heat of vaporization: ¢Hsub = ¢Hfus + ¢Hvap. Figure 11.24 Increases, because the molecules have more kinetic energy as the temperature increases and can escape more easily Figure 11.25 All liquids including ethylene glycol reach their normal boiling point when their vapor pressure is equal to atmospheric pressure, 760 torr. Figure 11.27 Freezing, because for most substances the solid phase is denser than the liquid phase and increasing the pressure will eventually drive a phase transition from the liquid to the solid state (provided the temperature is below the critical temperature)
CHAPTER 12 Figure 12.13 A hexagonal lattice
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Figure 12.15 The solvent is the majority component and the solute the minority component. Therefore, there will be more solvent atoms than solute atoms. Figure 12.17 The samarium atoms sit on the corners of the unit cell so there is only 8 * (1/8) = 1 Sm atom per unit cell. Eight of the nine cobalt atoms sit on faces of the unit cell, and the other sits in the middle of the unit cell so there are 8 * (1/2) + 1 = 5 Co atoms per unit cell. Figure 12.19 The atoms are randomly arranged in red gold, which is a substitutional alloy. Purple gold is an intermetallic compound in which the atoms are arranged in a specific ordered pattern. Figure 12.20 By drawing Lewis structures you can show that there are three (chlorine), two (sulfur), one (phosphorus), and zero (silicon) nonbonding electron pairs per atom. Figure 12.22 In the fourth period, vanadium and chromium have very similar melting points. Molybdenum and tungsten have the highest melting points in the fifth and sixth periods, respectively. All of these elements are located near the middle of the period where the bonding orbitals are mostly filled and the antibonding orbitals mostly empty. Figure 12.23 The molecular orbitals become more closely spaced in energy. Figure 12.24 Potassium has only one valence electron per atom. If we fill the 4s band halfway probably a small amount of electron density will leak over and start to fill the 3d orbitals as well. The 4p orbitals should be empty. Figure 12.25 Ionic substances cleave because the nearest neighbor interactions switch from attractive to repulsive if the atoms slide so that ions of like charge (cation–cation and anion–anion) touch each other. Metals don’t cleave because the atoms are attracted to all other atoms in the crystal through metallic bonding. Figure 12.26 No, ions of like charge do not touch in an ionic compound because they are repelled from one another. In an ionic compound the cations touch the anions. Figure 12.28 In NaF there are four Na⫹ ions (12 * 1/4) and four F- ions (8 * 1/8 + 6 * 1/2) per unit cell. In MgF2 there are two Mg2⫹ ions (8 * 1/8 + 1) and four F- ions (4 * 1/2 + 2) per unit cell. In ScF3 there is one Sc3⫹ ion (8 * 1/8) and three F- ions (12 * 1/4) per unit cell. Figure 12.29 The intermolecular forces are stronger in toluene, as shown by its higher boiling point. The molecules pack more efficiently in benzene, which explains its higher melting point, even though the intermolecular forces are weaker. Figure 12.44 Decrease. As the quantum dots get smaller, the band gap increases and the emitted light shifts to shorter wavelength. Figure 12.47 Each carbon atom in C60 is bonded to three neighboring carbon atoms through covalent bonds. Thus, the bonding is more like graphite, where carbon atoms also bond to three neighbors, than diamond, where carbon atoms bond to four neighbors.
CHAPTER 13 Figure 13.2 Opposite charges attract. The electron-rich O atom of the H2O molecule, which is the negative end of the dipole, is attracted to the positive Na⫹ ion. Figure 13.3 The negative end of the water dipole (the O) is attracted to the positive Na⫹ ion, whereas the positive end of the dipole (the H) is attracted to the negative Cl⫺ ion. Figure 13.4 For exothermic solution processes the magnitude of ¢Hmix will be larger than the magnitude of ¢Hsolute ⫹ ¢Hsolvent Figure 13.8 The dissolving of the crystal and the crystallization by which ions in solution become reattached to the solid Figure 13.9 If the solution wasn’t supersaturated, solute would not crystallize from it.
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Answers to Go Figure
Figure 13.14 If the partial pressure of a gas over a solution is doubled, the concentration of gas in the solution would double. Figure 13.15 The slopes increase as the molecular weight increases. The larger the molecular weight, the greater the polarizability of the gas molecules, leading to greater intermolecular attractive forces between gas molecules and water molecules. Figure 13.18 Looking at where the solubility curves for KCl and NaCl intersect the 80 °C line, we see that the solubility of KCl is about 51 g/100 g H2O, whereas NaCl has a solubility of about 39 g/100 g H2O. Thus, KCl is more soluble than NaCl at this temperature. Figure 13.19 N2 has the same molecular weight as CO but is nonpolar, so we can predict that its curve will be just below that of CO. Figure 13.25 The water will move through the semipermeable membrane toward the more concentrated solution. Thus, the liquid level in the left arm will increase. Figure 13.26 Water will move toward the more concentrated solute solution, which is inside the red blood cells, causing them to undergo hemolysis. Figure 13.30 The two negatively charged groups both have the composition ¬ CO2 - . Figure 13.32 Recall the rule that likes dissolve likes. The oil drop is composed of nonpolar molecules, which interact with the nonpolar part of the stearate ion with dispersion forces.
CHAPTER 14 Figure 14.3 B Figure 14.4 It decreases. Figure 14.8 The reaction is first order in CH3NC. Figure 14.10 At early times in the reaction; both graphs look linear close to t = 0. Figure 14.17 The energy needed to overcome the energy barrier (the activation energy) looks about twice as large as the overall energy change for the reaction. Figure 14.23 For the blue curve: The transition states are at the top of the peaks (2) and the intermediate is in the “valley” between the two peaks. For the red curve: The top of the peak is the transition state; no intermediates are shown. Figure 14.27 Substrate; if products bound tightly, they would not leave and the active site would not be free.
CHAPTER 15 Figure 15.1 The color in the tube stops changing. Figure 15.2 No Figure 15.6 The boxes would be approximately the same size. Figure 15.7 It will be lower; some CO2 has to react with CaO to make some CaCO3. Figure 15.9 500 atm and 400 °C Figure 15.10 Nitrogen (and some of the added hydrogen) is converted into ammonia. Figure 15.14 About two to three times faster, based on the graph Figure 15.15 About 5 * 10-4
CHAPTER 16 Figure 16.2 Hydrogen bonds Figure 16.3 O2 - (aq) + H2O(l) ¡ 2 OH - (aq) Figure 16.7 Phenolphthalein changes from colorless, for pH values less than 8, to pink for pH values greater than 10. A pink color indicates pH 7 10. Figure 16.8 Bromothymol blue would be most suitable because it changes pH over a range that brackets pH = 7. Methyl red is not sensitive to
pH changes when pH 7 6, while phenolphthalein is not sensitive to pH changes when pH 6 8, so neither changes color at pH = 7. Figure 16.11 Yes. The equilibrium of interest is H3CCOOH Δ H⫹ ⫹ H3CCOO⫺. If the percent dissociation remained constant as the acid concentration increased, the concentration of all three species would increase at the same rate. However, because there are two products and only one reactant, the total concentration of products would increase faster than the concentration of reactants. To offset this effect the percent dissociation decreases as the acid concentration increases. Figure 16.12 The acidic hydrogens belong to carboxlyate ( ¬ COOH) groups, whereas the fourth proton bound to oxygen is part of a hydroxyl ( ¬ OH) group. In organic acids, like citric acid, the acidic protons are almost always part of a carboxylate group. Figure 16.13 The nitrogen atom in hydroxylamine accepts a proton to form NH3OH⫹. As a general rule, nonbonding electron pairs on nitrogen atoms are more basic than nonbonding electron pairs on oxygen atoms.
CHAPTER 17 Figure 17.6 The pH will increase on addition of the base. Figure 17.7 25.00 mL. The number of moles of added base needed to reach the equivalence point remains the same. Therefore, by doubling the concentration of added base the volume needed to reach the equivalence point is halved. Figure 17.9 The volume of base needed to reach the equivalence point would not change because this quantity does not depend on the strength of the acid. However, the pH at the equivalence point, which is greater than 7 for a weak acid–strong base titration, would decrease to 7 because hydrochloric acid is a strong acid. Figure 17.11 The pH at the equivalence point increases (becomes more basic) as the acid becomes weaker. The volume of added base needed to reach the equivalence point remains unchanged. Figure 17.13 Yes. Any indicator that changes color between pH = 3 and pH = 11 could be used for a strong acid–strong base titration. Methyl red changes color between pH values of approximately 4 and 6. Figure 17.22 ZnS and CuS would both precipitate on addition of H2S, preventing separation of the two ions. Figure 17.23 Yes. CuS would precipitate in step 2 on addition of H2S to an acidic solution, while the Zn2⫹ ions remained in solution.
CHAPTER 18 Figure 18.1 About 85 km Figure 18.3 The atmosphere absorbs a significant fraction of solar radiation. Figure 18.4 The peak value is about 5 * 1012 molecules per cm3. If we use Avogradro’s number to convert molecules to moles, and the conversion factor of 1000 cm3 = 1000 mL = 1 L, we find that the concentration of ozone at the peak is 8 * 10 - 9 mole/L. Figure 18.16 This is ambiguous; both temperature and salinity vary with density in similar ways; but temperature seems to parallel density better. Temperature decreases down to 1000 m, then remains relatively constant; density increases down to 1000 m, and then remains relatively constant. Figure 18.17 The depth of the aquifer; the nature of the intervening layers (how porous or dense they are) Figure 18.19 Water is the chemical species that is crossing the membrane, not the ions.
CHAPTER 19 Figure 19.1 Yes, the potential energy of the eggs decreases as they fall. Figure 19.2 Because the final volume would be less than twice the volume of Flask A, the final pressure would be greater than 0.5 atm.
Answers to Go Figure Figure 19.3 The freezing of liquid water to ice is exothermic. Figure 19.4 To be truly reversible, the temperature change dT must be infinitesimally small. Figure 19.8 There are two other independent rotational motions of the H2O molecule:
Figure 19.9 Ice, because it is the phase in which the molecules are held most rigidly Figure 19.11 The decrease in the number of molecules due to the formation of new bonds. Figure 19.13 During a phase change, the temperature remains constant but the entropy change can be large. Figure 19.14 Based on the three molecules shown, the addition of each C increases S° by 40–45 J/mol-K. Based on this observation, we would predict that S°(C4H10) would 310–315 J/mol-K. Appendix C confirms that this is a good prediction: S°(C4H10) = 310.0 J/mol-K. Figure 19.16 Spontaneous Figure 19.17 If we plot progress of the reaction versus free energy, equilibrium is at a minimum point in free energy, as shown in the figure. In that sense, the reaction runs “downhill” until it reaches that minimum point.
CHAPTER 20 Figure 20.1 (a) The bubbling is caused by the hydrogen gas formed in the reaction. (b) The reaction is exothermic, and the heat causes the formation of steam. Figure 20.2 The permanganate, MnO4 - , is reduced, as the half-reactions in the text show. Figure 20.3 The blue color is due to Cu2⫹(aq). As this ion is reduced, forming Cu(s), the blue color fades. Figure 20.4 The Zn is oxidized and, therefore, serves as the anode of the cell. Figure 20.5 The electrical balance is maintained in two ways: Anions migrate into the half-cell, and cations migrate out. Figure 20.9 As the cell operates, H⫹ is reduced to H2 in the cathode half-cell. As H⫹ is depleted, the positive Na⫹ ions are drawn into the half-cell to maintain electrical balance in the solution. Figure 20.10 The reduction reaction occurs at the cathode. The substance that is reduced most easily is the one with the larger standard ° . reduction potential, E red Figure 20.12 Oxidation is the loss of electrons. An oxidizing agent causes another substance to lose electrons by gaining them itself. A strong oxidizing agent readily gains electrons, meaning that it is easily reduced. Figure 20.14 The variable n is the number of moles of electrons transferred in the process. Figure 20.15 The Ni2⫹(aq) and the cations in the salt bridge migrate toward the cathode. The NO3⫺(aq) and the anions in the salt bridge migrate toward the anode. Figure 20.19 The cathode consists of PbO2(s). Because each oxygen has an oxidation state of -2, lead must have an oxidation state of + 4 in this compound. Figure 20.22 The oxidizing agent of O2(g) from the air
CHAPTER 21 Figure 21.2 From Figure 21.2 we see that the belt of stability for a nucleus containing 70 protons lies at approximately 102 neutrons.
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Figure 21.4 Only three of the elements with an even number of protons have fewer than three isotopes: He, Be, and C. Note that these three elements are the lightest elements that have an even atomic number. Because they are so light, any change in the number of neutrons will change the neutron/proton ratio significantly. This helps to explain why they do not have more stable isotopes. None of the elements in Figure 21.4 that have an odd number of protons have more than two stable isotopes. Figure 21.6 6.25 g. After one half-life, the amount of the radioactive material will have dropped to 25.0 g. After two half-lives, it will have dropped to 12.5 g. After three half-lives, it will have dropped to 6.25 g. Figure 21.19 Because large quantities of water are needed to condense the secondary coolant once it passes through the turbine Figure 21.21 The United States has the most reactors in operation. China has the most reactors under construction. France generates the largest percentage of its electricity from nuclear power. Figure 21.24 Alpha rays are less dangerous when outside the body because they cannot penetrate the skin. However, once inside the body they can do great harm to any cells they come in contact with.
CHAPTER 22 Figure 22.5 Beaker on the right is warmer. Figure 22.6 HF is the most stable, SbH3 the least stable. Figure 22.8 More soluble in CCl4—the colors are deeper. Figure 22.9 CF2 Figure 22.10 Redox reactions: The halides are being oxidized. Figure 22.14 No Figure 22.16 Based on this structure—yes, it would have a dipole moment. In fact, if you look it up, hydrogen peroxide’s dipole moment is larger than water’s! Figure 22.20 They have been converted into water. Figure 22.21 Formally they could both be +2. If we consider that the central sulfur is like SO42⫺, however, then the central sulfur would be +6, like SO42⫺, and then the terminal sulfur would be -2. Figure 22.22 Nitrite Figure 22.23 Longer Figure 22.26 The NO double bond Figure 22.28 In P4O6 the electron domains about the P atoms are trigonal pyramidal; in P4O10 the electron domains about the P atoms are tetrahedral. Figure 22.33 The minimum temperature should be the melting point of silicon; the temperature of the heating coil should not be so high that the silicon rod starts to melt outside the zone of the heating coil.
CHAPTER 23 Figure 23.3 Zn (it is colorless) Figure 23.4 The increase parallels the linear increase in valence electron count. Figure 23.5 All the electron spins would align with the direction of the magnetic field. Figure 23.9 109.5 degrees for the tetrahedral Zn complex; 90 degrees for the square-planar Pt complex Figure 23.13 4 for both (assuming no other ligands come in to bind) Figure 23.15 In the same place as O2 Figure 23.16 The peak with a maximum at 650 nm, the longest wavelength and lowest energy Figure 23.21 The cis one Figure 23.24 Larger, since ammonia can displace water Figure 23.26 The peak would stay in the same position in terms of wavelength, but its absorbance would decrease.
A-46
Answers to Go Figure
Figure 23.28 dx2-y2 and dz2 Figure 23.29 Convert the wavelength of light, 495 nm, into energy in joules using E = hc/l. Figure 23.30 It would be to the right of the “yellow” member of the series, but the energy gap between filled and empty d orbitals would be even larger than that of the “yellow” one. Figure 23.34 That orbital has the lobes that point directly at the ligands.
CHAPTER 24 Figure 24.1 Tetrahedral Figure 24.2 The OH group is polar whereas the CH3 group is nonpolar. Hence, adding CH3 will (a) reduce the substance’s solubility in polar solvents and (b) increase its solubility in nonpolar solvents. Figure 24.5 CnH2n, because there are no CH3 groups, each carbon has two hydrogens.
Figure 24.7 Just one Figure 24.9 Intermediates are minima and transition states are maxima on energy profiles. Figure 24.14 Both lactic acid and citric acid Figure 24.15 No, because there are not four different groups around any carbon Figure 24.18 Those labeled “basic amino acids,” which have basic side groups that are protonated at pH 7 Figure 24.25 The long hydrocarbon chains, which are nonpolar Figure 24.27 The polar parts of the phospholipids seek to interact with water whereas the nonpolar parts seek to interact with other nonpolar substances and to avoid water. Figure 24.29 Negative charge because of charge on phosphate groups Figure 24.31 GC because each base has three hydrogen bonding sites, whereas there are only two in AT
GLOSSARY absolute zero The lowest attainable temperature; 0 K on the Kelvin scale and -273.15 °C on the Celsius scale. (Section 1.4)
alkanes Compounds of carbon and hydrogen containing only carbon–carbon single bonds. (Sections 2.9 and 24.2)
absorption spectrum A pattern of variation in the amount of light absorbed by a sample as a function of wavelength. (Section 23.5)
alkenes Hydrocarbons containing one or more carbon–carbon double bonds. (Section 24.2)
accuracy A measure of how closely individual measurements agree with the correct value. (Section 1.5) acid A substance that is able to donate a H + ion (a proton) and, hence, increases the concentration of H + (aq) when it dissolves in water. (Section 4.3) acid-dissociation constant (Ka) An equilibrium constant that expresses the extent to which an acid transfers a proton to solvent water. (Section 16.6) acidic anhydride (acidic oxide) An oxide that forms an acid when added to water; soluble nonmetal oxides are acidic anhydrides. (Section 22.5) acidic oxide (acidic anhydride) An oxide that either reacts with a base to form a salt or with water to form an acid. (Section 22.5) acid rain Rainwater that has become excessively acidic because of absorption of pollutant oxides, notably SO3, produced by human activities. (Section 18.2) actinide element Element in which the 5f orbitals are only partially occupied. (Section 6.8) activated complex (transition state) The particular arrangement of atoms found at the top of the potential-energy barrier as a reaction proceeds from reactants to products. (Section 14.5) activation energy (Ea) The minimum energy needed for reaction; the height of the energy barrier to formation of products. (Section 14.5) active site Specific site on a heterogeneous catalyst or an enzyme where catalysis occurs. (Section 14.7) activity The decay rate of a radioactive material, generally expressed as the number of disintegrations per unit time. (Section 21.4) activity series A list of metals in order of decreasing ease of oxidation. (Section 4.4) addition polymerization Polymerization that occurs through coupling of monomers with one another, with no other products formed in the reaction. (Section 12.8) addition reaction A reaction in which a reagent adds to the two carbon atoms of a carbon–carbon multiple bond. (Section 24.3) adsorption The binding of molecules to a surface. (Section 14.7) alcohol An organic compound obtained by substituting a hydroxyl group (¬OH) for a hydrogen on a hydrocarbon. (Sections 2.9 and 24.4) aldehyde An organic compound that contains a carbonyl group (C “ O) to which at least one hydrogen atom is attached. (Section 24.4) alkali metals Members of group 1A in the periodic table. (Section 7.7) alkaline earth metals Members of group 2A in the periodic table. (Section 7.7)
alkyl group A group that is formed by removing a hydrogen atom from an alkane. (Section 25.3) alkynes Hydrocarbons containing one or more carbon–carbon triple bonds. (Section 24.2) alloy A substance that has the characteristic properties of a metal and contains more than one element. Often there is one principal metallic component, with other elements present in smaller amounts. Alloys may be homogeneous or heterogeneous. (Section 12.3) alpha decay A type of radioactive decay in which an atomic nucleus emits an alpha particle and thereby transforms (or “decays”) into an atom with a mass number 4 less and atomic number 2 less. (Section 21.1) alpha (a) helix A protein structure in which the protein is coiled in the form of a helix with hydrogen bonds between C “ O and N ¬ H groups on adjacent turns. (Section 24.7) alpha particles Particles that are identical to helium-4 nuclei, consisting of two protons and two neutrons, symbol 42He or 42a. (Section 21.1) amide An organic compound that has an NR2 group attached to a carbonyl. (Section 24.4) amine A compound that has the general formula R3N, where R may be H or a hydrocarbon group. (Section 16.7) amino acid A carboxylic acid that contains an amino (¬NH2) group attached to the carbon atom adjacent to the carboxylic acid (¬COOH) functional group. (Section 24.7) amorphous solid A solid whose molecular arrangement lacks the regularly repeating longrange pattern of a crystal. (Section 12.2) amphiprotic Refers to the capacity of a substance to either add or lose a proton (H + ). (Section 16.2) amphoteric oxides and hydroxides Oxides and hydroxides that are only slightly soluble in water but that dissolve in either acidic or basic solutions. (Section 17.5) angstrom A common non-SI unit of length, denoted ⴰ Å, that is used to measure atomic dimensions: 1〈 = 10-10 m. (Section 2.3) anion
A negatively charged ion. (Section 2.7)
anode An electrode at which oxidation occurs. (Section 20.3)
aqueous solution A solution in which water is the solvent. (Chapter 4: Introduction) aromatic hydrocarbons Hydrocarbon compounds that contain a planar, cyclic arrangement of carbon atoms linked by both s and delocalized p bonds. (Section 24.2) Arrhenius equation An equation that relates the rate constant for a reaction to the frequency factor, A, the activation energy, Ea, and the temperature, T: k = Ae -Ea>RT. In its logarithmic form it is written ln k = -Ea>RT + ln A. (Section 14.5) atmosphere (atm) A unit of pressure equal to 760 torr; 1 atm = 101.325 kPa. (Section 10.2) atom The smallest representative particle of an element. (Sections 1.1 and 2.1) atomic mass unit (amu) A unit based on the value of exactly 12 amu for the mass of the isotope of carbon that has six protons and six neutrons in the nucleus. (Sections 2.3 and 3.3) atomic number The number of protons in the nucleus of an atom of an element. (Section 2.3) atomic radius An estimate of the size of an atom. See bonding atomic radius. (Section 7.3) atomic weight The average mass of the atoms of an element in atomic mass units (amu); it is numerically equal to the mass in grams of one mole of the element. (Section 2.4) autoionization The process whereby water spontaneously forms low concentrations of H + (aq) and OH-(aq) ions by proton transfer from one water molecule to another. (Section 16.3) Avogadro’s hypothesis A statement that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. (Section 10.3) Avogadro’s law A statement that the volume of a gas maintained at constant temperature and pressure is directly proportional to the number of moles of the gas. (Section 10.3) Avogadro’s number (NA) The number of 12C atoms in exactly 12 g of 12C; it equals 6.022 * 1023 mol-1. (Section 3.4) band An array of closely spaced molecular orbitals occupying a discrete range of energy. (Section 12.4) band gap The energy gap between a fully occupied band called a valence band and an empty band called the conduction band. (Section 12.7) band structure The electronic structure of a solid, defining the allowed ranges of energy for electrons in a solid. (Section 12.7)
antibonding molecular orbital A molecular orbital in which electron density is concentrated outside the region between the two nuclei of bonded atoms. Such orbitals, designated as s* or p*, are less stable (of higher energy) than bonding molecular orbitals. (Section 9.7)
bar A unit of pressure equal to 105 Pa. (Section 10.2)
antiferromagnetism A form of magnetism in which unpaired electron spins on adjacent sites point in opposite directions and cancel each other’s effects. (Section 23.1)
base-dissociation constant (Kb) An equilibrium constant that expresses the extent to which a base reacts with solvent water, accepting a proton and forming OH-(aq). (Section 16.7)
base A substance that is an H + acceptor; a base produces an excess of OH-(aq) ions when it dissolves in water. (Section 4.3)
G-1
G-2
GLOSSARY
basic anhydride (basic oxide) An oxide that forms a base when added to water; soluble metal oxides are basic anhydrides. (Section 22.5)
of antibonding electron pairs: bond order = (number of bonding electrons - number of antibonding electrons)> 2. (Section 9.7)
basic oxide (basic anhydride) An oxide that either reacts with water to form a base or reacts with an acid to form a salt and water. (Section 22.5)
bond polarity A measure of the degree to which the electrons are shared unequally between two atoms in a chemical bond. (Section 8.4)
battery A self-contained electrochemical power source that contains one or more voltaic cells. (Section 20.7)
boranes Covalent hydrides of boron. (Section 22.11)
becquerel The SI unit of radioactivity. It corresponds to one nuclear disintegration per second. (Section 21.4)
Born–Haber cycle A thermodynamic cycle based on Hess’s law that relates the lattice energy of an ionic substance to its enthalpy of formation and to other measurable quantities. (Section 8.2)
Beer’s law The light absorbed by a substance (A) equals the product of its extinction coefficient (e), the path length through which the light passes (b), and the molar concentration of the substance (c): A = ebc.(Section 14.2)
Boyle’s law A law stating that at constant temperature, the product of the volume and pressure of a given amount of gas is a constant. (Section 10.3)
beta emission A nuclear decay process where a beta particle is emitted from the nucleus; also called beta decay. (Section 21.1)
Brønsted–Lowry base A substance (molecule or ion) that acts as a proton acceptor. (Section 16.2)
Brønsted–Lowry acid A substance (molecule or ion) that acts as a proton donor. (Section 16.2)
beta particles Energetic electrons emitted from the nucleus, symbol -10e. (Section 21.1)
buffer capacity The amount of acid or base a buffer can neutralize before the pH begins to change appreciably. (Section 17.2)
beta sheet A structural form of protein in which two strands of amino acids are hydrogenbonded together in a zipperlike configuration. (Section 24.7)
buffered solution (buffer) A solution that undergoes a limited change in pH upon addition of a small amount of acid or base. (Section 17.2)
bidentate ligand A ligand in which two linked coordinating atoms are bound to a metal. (Section 23.3) bimolecular reaction An elementary reaction that involves two molecules. (Section 14.6) biochemistry The study of the chemistry of living systems. (Chapter 24: Introduction) biodegradable Organic material that bacteria are able to oxidize. (Section 18.4) body-centered lattice A crystal lattice in which the lattice points are located at the center and corners of each unit cell. (Section 12.2) bomb calorimeter A device for measuring the heat evolved in the combustion of a substance under constant-volume conditions. (Section 5.5) bond angles The angles made by the lines joining the nuclei of the atoms in a molecule. (Section 9.1) bond dipole The dipole moment that is due to unequal electron sharing between two atoms in a covalent bond. (Section 9.3) bond enthalpy The enthalpy change, ¢H, required to break a particular bond when the substance is in the gas phase. (Section 8.8) bonding atomic radius The radius of an atom as defined by the distances separating it from other atoms to which it is chemically bonded. (Section 7.3) bonding molecular orbital A molecular orbital in which the electron density is concentrated in the internuclear region. The energy of a bonding molecular orbital is lower than the energy of the separate atomic orbitals from which it forms. (Section 9.7) bonding pair In a Lewis structure a pair of electrons that is shared by two atoms. (Section 9.2)
calcination The heating of an ore to bring about its decomposition and the elimination of a volatile product. For example, a carbonate ore might be calcined to drive off CO2. (Section 23.2)
a voltaic cell. This can be achieved by attaching a more easily oxidized metal, which serves as an anode, to the metal to be protected. (Section 20.8) cation
A positively charged ion. (Section 2.7)
cell potential The potential difference between the cathode and anode in an electrochemical cell; it is measured in volts: 1 V = 1 J>C. Also called electromotive force. (Section 20.4) cellulose A polysaccharide of glucose; it is the major structural element in plant matter. (Section 24.8) Celsius scale A temperature scale on which water freezes at 0° and boils at 100° at sea level. (Section 1.4) chain reaction A series of reactions in which one reaction initiates the next. (Section 21.7) changes of state Transformations of matter from one state to a different one, for example, from a gas to a liquid. (Section 1.3) charcoal A form of carbon produced when wood is heated strongly in a deficiency of air. (Section 22.9) Charles’s law A law stating that at constant pressure, the volume of a given quantity of gas is proportional to absolute temperature. (Section 10.3) chelate effect The generally larger formation constants for polydentate ligands as compared with the corresponding monodentate ligands. (Section 23.3) chelating agent A polydentate ligand that is capable of occupying two or more sites in the coordination sphere. (Section 23.3)
calorie A unit of energy, it is the amount of energy needed to raise the temperature of 1 g of water by 1 °C from 14.5 °C to 15.5 °C. A related unit is the joule: 1 cal = 4.184 J. (Section 5.1)
chemical bond A strong attractive force that exists between atoms in a molecule. (Section 8.1)
calorimeter An apparatus that measures the heat released or absorbed in a chemical or physical process. (Section 5.5)
chemical changes Processes in which one or more substances are converted into other substances; also called chemical reactions. (Section 1.3)
calorimetry The experimental measurement of heat produced in chemical and physical processes. (Section 5.5)
chemical equation A representation of a chemical reaction using the chemical formulas of the reactants and products; a balanced chemical equation contains equal numbers of atoms of each element on both sides of the equation. (Section 3.1)
capillary action The process by which a liquid rises in a tube because of a combination of adhesion to the walls of the tube and cohesion between liquid particles. (Section 11.3) carbide A binary compound of carbon with a metal or metalloid. (Section 22.9) carbohydrates A class of substances formed from polyhydroxy aldehydes or ketones. (Section 24.8) carbon black A microcrystalline form of carbon. (Section 22.9) carbonyl group The C “ O double bond, a characteristic feature of several organic functional groups, such as ketones and aldehydes. (Section 24.4) carboxylic acid A compound that contains the ¬ COOH functional group. (Sections 16.10 and 24.4) catalyst A substance that changes the speed of a chemical reaction without itself undergoing a permanent chemical change in the process. (Section 14.7) cathode An electrode at which reduction occurs. (Section 20.3)
bond length The distance between the centers of two bonded atoms. (Section 8.3)
cathode rays Streams of electrons that are produced when a high voltage is applied to electrodes in an evacuated tube. (Section 2.2)
bond order The number of bonding electron pairs shared between two atoms, minus the number
cathodic protection A means of protecting a metal against corrosion by making it the cathode in
chemical equilibrium A state of dynamic balance in which the rate of formation of the products of a reaction from the reactants equals the rate of formation of the reactants from the products; at equilibrium the concentrations of the reactants and products remain constant. (Section 4.1; Chapter 15: Introduction) chemical formula A notation that uses chemical symbols with numerical subscripts to convey the relative proportions of atoms of the different elements in a substance. (Section 2.6) chemical kinetics The area of chemistry concerned with the speeds, or rates, at which chemical reactions occur. (Chapter 14: Introduction) chemical nomenclature The rules used in naming substances. (Section 2.8) chemical properties Properties that describe a substance’s composition and its reactivity; how the substance reacts or changes into other substances. (Section 1.3) chemical reactions Processes in which one or more substances are converted into other substances; also called chemical changes. (Section 1.3) chemistry The scientific discipline that studies the composition, properties, and transformations of matter. (Chapter 1: Introduction)
GLOSSARY chiral A term describing a molecule or an ion that cannot be superimposed on its mirror image. (Sections 23.4 and 24.5) chlorofluorocarbons Compounds composed entirely of chlorine, fluorine, and carbon. (Section 18.3) chlorophyll A plant pigment that plays a major role in conversion of solar energy to chemical energy in photosynthesis. (Section 23.3) cholesteric liquid crystalline phase A liquid crystal formed from flat, disc-shaped molecules that align through a stacking of the molecular discs. (Section 11.7) coal A naturally occurring solid containing hydrocarbons of high molecular weight, as well as compounds containing sulfur, oxygen, and nitrogen. (Section 5.8) colligative property A property of a solvent (vapor-pressure lowering, freezing-point lowering, boiling-point elevation, osmotic pressure) that depends on the total concentration of solute particles present. (Section 13.5) collision model A model of reaction rates based on the idea that molecules must collide to react; it explains the factors influencing reaction rates in terms of the frequency of collisions, the number of collisions with energies exceeding the activation energy, and the probability that the collisions occur with suitable orientations. (Section 14.5) colloids (colloidal dispersions) Mixtures containing particles larger than normal solutes but small enough to remain suspended in the dispersing medium. (Section 13.6) combination reaction A chemical reaction in which two or more substances combine to form a single product. (Section 3.2) combustion reaction A chemical reaction that proceeds with evolution of heat and usually also a flame; most combustion involves reaction with oxygen, as in the burning of a match. (Section 3.2) common-ion effect A shift of an equilibrium induced by an ion common to the equilibrium. For example, added Na2SO4 decreases the solubility of the slightly soluble salt BaSO4, or added NaF decreases the percent ionization of HF. (Section 17.1) complementary colors Colors that, when mixed in proper proportions, appear white or colorless. (Section 23.5) complete ionic equation A chemical equation in which dissolved strong electrolytes (such as dissolved ionic compounds) are written as separate ions. (Section 4.2) complex ion (complex) An assembly of a metal ion and the Lewis bases (ligands) bonded to it. (Section 17.5) compound A substance composed of two or more elements united chemically in definite proportions. (Section 1.2) compound semiconductor A semiconducting material formed from two or more elements. (Section 12.7) concentration The quantity of solute present in a given quantity of solvent or solution. (Section 4.5) concentration cell A voltaic cell containing the same electrolyte and the same electrode materials in both the anode and cathode compartments. The emf of the cell is derived from a difference in the
concentrations of the same electrolyte solutions in the compartments. (Section 20.6) condensation polymerization Polymerization in which molecules are joined together through condensation reactions. (Section 12.8) condensation reaction A chemical reaction in which a small molecule (such as a molecule of water) is split out from between two reacting molecules. (Sections 12.6 and 22.8) conduction band A band of molecular orbitals lying higher in energy than the occupied valence band and distinctly separated from it. (Section 12.7) conjugate acid A substance formed by addition of a proton to a Brønsted–Lowry base. (Section 16.2) conjugate acid–base pair An acid and a base, such as H2O and OH- , that differ only in the presence or absence of a proton. (Section 16.2) conjugate base A substance formed by the loss of a proton from a Brønsted–Lowry acid. (Section 16.2) continuous spectrum A spectrum that contains radiation distributed over all wavelengths. (Section 6.3) conversion factor A ratio relating the same quantity in two systems of units that is used to convert the units of measurement. (Section 1.6) coordination compound A compound containing a metal ion bonded to a group of surrounding molecules or ions that act as ligands. (Section 23.2) coordination number The number of adjacent atoms to which an atom is directly bonded. In a complex the coordination number of the metal ion is the number of donor atoms to which it is bonded. (Sections 12.37 and 24.2) coordination sphere The metal ion and its surrounding ligands. (Section 23.2) coordination-sphere isomers Structural isomers of coordination compounds in which the ligands within the coordination sphere differ. (Section 23.4) copolymer A complex polymer resulting from the polymerization of two or more chemically different monomers. (Section 12.8) core electrons The electrons that are not in the outermost shell of an atom. (Section 6.8) corrosion The process by which a metal is oxidized by substances in its environment. (Section 20.8) covalent bond A bond formed between two or more atoms by a sharing of electrons. (Section 8.1) covalent-network solids Solids in which the units that make up the three-dimensional network are joined by covalent bonds. (Section 12.1) critical mass The amount of fissionable material necessary to maintain a nuclear chain reaction. (Section 21.7) critical pressure The pressure at which a gas at its critical temperature is converted to a liquid state. (Section 11.4) critical temperature The highest temperature at which it is possible to convert the gaseous form of a substance to a liquid. The critical temperature increases with an increase in the magnitude of intermolecular forces. (Section 11.4)
G-3
crystal-field theory A theory that accounts for the colors and the magnetic and other properties of transition-metal complexes in terms of the splitting of the energies of metal ion d orbitals by the electrostatic interaction with the ligands. (Section 23.6) crystal lattice An imaginary network of points on which the repeating motif of a solid may be imagined to be laid down so that the structure of the crystal is obtained. The motif may be a single atom or a group of atoms. Each lattice point represents an identical environment in the crystal. (Section 12.2) crystalline solid (crystal) A solid whose internal arrangement of atoms, molecules, or ions possesses a regularly repeating pattern in any direction through the solid. (Section 12.2) crystallization The process in which molecules, ions, or atoms come together to form a crystalline solid. (Section 13.2) cubic close packing A crystal structure where the atoms are packed together as close as possible, and the close-packed layers of atoms adopt a threelayer repeating pattern that leads to a face-centered cubic unit cell. (Section 12.3) curie A measure of radioactivity: 1 curie = 3.7 * 1010 nuclear disintegrations per second. (Section 21.4) cycloalkanes Saturated hydrocarbons of general formula CnH2n in which the carbon atoms form a closed ring. (Section 24.2) Dalton’s law of partial pressures A law stating that the total pressure of a mixture of gases is the sum of the pressures that each gas would exert if it were present alone. (Section 10.6) d-d transition The transition of an electron in a transition-metal compound from a lower-energy d orbital to a higher-energy d orbital. (Section 23.6) decomposition reaction A chemical reaction in which a single compound reacts to give two or more products. (Section 3.2) degenerate A situation in which two or more orbitals have the same energy. (Section 6.7) delocalized electrons Electrons that are spread over a number of atoms in a molecule or a crystal rather than localized on a single atom or a pair of atoms. (Section 9.6) density The ratio of an object’s mass to its volume. (Section 1.4) deoxyribonucleic acid (DNA) A polynucleotide in which the sugar component is deoxyribose. (Section 24.10) desalination The removal of salts from seawater, brine, or brackish water to make it fit for human consumption. (Section 18.4) deuterium The isotope of hydrogen whose nucleus contains a proton and a neutron: 21H. (Section 22.2) dextrorotatory, or merely dextro or d A term used to label a chiral molecule that rotates the plane of polarization of plane-polarized light to the right (clockwise). (Section 23.4) diamagnetism A type of magnetism that causes a substance with no unpaired electrons to be weakly repelled from a magnetic field. (Section 9.8) diatomic molecule A molecule composed of only two atoms. (Section 2.6)
G-4
GLOSSARY
diffusion The spreading of one substance through a space occupied by one or more other substances. (Section 10.8) dilution The process of preparing a less concentrated solution from a more concentrated one by adding solvent. (Section 4.5) dimensional analysis A method of problem solving in which units are carried through all calculations. Dimensional analysis ensures that the final answer of a calculation has the desired units. (Section 1.6) dipole A molecule with one end having a partial negative charge and the other end having a partial positive charge; a polar molecule. (Section 8.4) dipole–dipole force A force that becomes significant when polar molecules come in close contact with one another. The force is attractive when the positive end of one polar molecule approaches the negative end of another. (Section 11.2)
electrolytic cell A device in which a nonspontaneous oxidation-reduction reaction is caused to occur by passage of current under a sufficient external electrical potential. (Section 20.9) electromagnetic radiation (radiant energy) A form of energy that has wave characteristics and that propagates through a vacuum at the characteristic speed of 3.00 * 108 m>s. (Section 6.1) electrometallurgy The use of electrolysis to reduce or refine metals. (Section 20.9) electromotive force (emf) A measure of the driving force, or electrical pressure, for the completion of an electrochemical reaction. Electromotive force is measured in volts: 1 V = 1 J>C. Also called the cell potential. (Section 20.4) electron A negatively charged subatomic particle found outside the atomic nucleus; it is a part of all atoms. An electron has a mass 1>1836 times that of a proton. (Section 2.3)
dipole moment A measure of the separation and magnitude of the positive and negative charges in polar molecules. (Section 8.4)
electron affinity The energy change that occurs when an electron is added to a gaseous atom or ion. (Section 7.5)
dispersion forces Intermolecular forces resulting from attractions between induced dipoles. Also called London dispersion forces. (Section 11.2)
electron capture A mode of radioactive decay in which an inner-shell orbital electron is captured by the nucleus. (Section 21.1)
displacement reaction A reaction in which an element reacts with a compound, displacing an element from it. (Section 4.4)
electron configuration The arrangement of electrons in the orbitals of an atom or molecule (Section 6.8)
donor atom The atom of a ligand that bonds to the metal. (Section 23.2)
electron density The probability of finding an electron at any particular point in an atom; this probability is equal to c2, the square of the wave function. Also called the probability density. (Section 6.5)
doping Incorporation of a hetero atom into a solid to change its electrical properties. For example, incorporation of P into Si. (Section 12.7) double bond A covalent bond involving two electron pairs. (Section 8.3) double helix The structure for DNA that involves the winding of two DNA polynucleotide chains together in a helical arrangement. The two strands of the double helix are complementary in that the organic bases on the two strands are paired for optimal hydrogen bond interaction. (Section 24.10) dynamic equilibrium A state of balance in which opposing processes occur at the same rate. (Section 11.5) effective nuclear charge The net positive charge experienced by an electron in a manyelectron atom; this charge is not the full nuclear charge because there is some shielding of the nucleus by the other electrons in the atom. (Section 7.2)
electron domain In the VSEPR model, a region about a central atom in which an electron pair is concentrated. (Section 9.2) electron-domain geometry The threedimensional arrangement of the electron domains around an atom according to the VSEPR model. (Section 9.2) electronegativity A measure of the ability of an atom that is bonded to another atom to attract electrons to itself. (Section 8.4) electronic charge The negative charge carried by an electron; it has a magnitude of 1.602 * 10-19 C. (Section 2.3) electronic structure The arrangement of electrons in an atom or molecule. (Chapter 6: Introduction)
overall chemical reaction consists of one or more elementary reactions or steps. (Section 14.6) empirical formula A chemical formula that shows the kinds of atoms and their relative numbers in a substance in the smallest possible wholenumber ratios. (Section 2.6) enantiomers Two mirror-image molecules of a chiral substance. The enantiomers are nonsuperimposable. (Section 23.4) endothermic process A process in which a system absorbs heat from its surroundings. (Section 5.2) energy The capacity to do work or to transfer heat. (Section 5.1) energy-level diagram A diagram that shows the energies of molecular orbitals relative to the atomic orbitals from which they are derived. Also called a molecular-orbital diagram. (Section 9.7) enthalpy A quantity defined by the relationship H = E + PV; the enthalpy change, ¢H, for a reaction that occurs at constant pressure is the heat evolved or absorbed in the reaction: ¢H = qp. (Section 5.3) enthalpy of formation The enthalpy change that accompanies the formation of a substance from the most stable forms of its component elements. (Section 5.7) enthalpy of reaction The enthalpy change associated with a chemical reaction. (Section 5.4) entropy A thermodynamic function associated with the number of different equivalent energy states or spatial arrangements in which a system may be found. It is a thermodynamic state function, which means that once we specify the conditions for a system—that is, the temperature, pressure, and so on—the entropy is defined. (Section 19.2) enzyme A protein molecule that acts to catalyze specific biochemical reactions. (Section 14.7) equilibrium constant The numerical value of the equilibrium-constant expression for a system at equilibrium. The equilibrium constant is most usually denoted by Kp for gas-phase systems or Kc for solution-phase systems. (Section 15.2) equilibrium-constant expression The expression that describes the relationship among the concentrations (or partial pressures) of the substances present in a system at equilibrium. The numerator is obtained by multiplying the concentrations of the substances on the product side of the equation, each raised to a power equal to its coefficient in the chemical equation. The denominator similarly contains the concentrations of the substances on the reactant side of the equation. (Section 15.2)
effusion The escape of a gas through an orifice or hole. (Section 10.8)
electron-sea model A model for the behavior of electrons in metals. (Section 12.4)
elastomer A material that can undergo a substantial change in shape via stretching, bending, or compression and return to its original shape upon release of the distorting force. (Section 12.6)
electron shell A collection of orbitals that have the same value of n. For example, the orbitals with n = 3 (the 3s, 3p, and 3d orbitals) comprise the third shell. (Section 6.5)
electrochemistry The branch of chemistry that deals with the relationships between electricity and chemical reactions. (Chapter 20: Introduction)
electron spin A property of the electron that makes it behave as though it were a tiny magnet. The electron behaves as if it were spinning on its axis; electron spin is quantized. (Section 6.7)
ester An organic compound that has an OR group attached to a carbonyl; it is the product of a reaction between a carboxylic acid and an alcohol. (Section 24.4)
element A substance consisting of atoms of the same atomic number. Historically defined as a substance that cannot be separated into simpler substances by chemical means. (Sections 1.1 and 1.2)
ether A compound in which two hydrocarbon groups are bonded to one oxygen. (Section 24.4)
electrolysis reaction A reaction in which a nonspontaneous redox reaction is brought about by the passage of current under a sufficient external electrical potential. The devices in which electrolysis reactions occur are called electrolytic cells. (Section 20.9) electrolyte A solute that produces ions in solution; an electrolytic solution conducts an electric current. (Section 4.1)
equivalence point The point in a titration at which the added solute reacts completely with the solute present in the solution. (Section 4.6)
elemental semiconductor A semiconducting material composed of just one element. (Section 12.7)
exchange (metathesis) reaction A reaction between compounds that when written as a molecular equation appears to involve the exchange of ions between the two reactants. (Section 4.2)
elementary reaction A process in a chemical reaction that occurs in a single event or step. An
excited state A higher energy state than the ground state. (Section 6.3)
GLOSSARY
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exothermic process A process in which a system releases heat to its surroundings. (Section 5.2)
fuel value The energy released when 1 g of a substance is combusted. (Section 5.8)
involved, for example, Zn2 + (aq) + 2 e - ¡ Zn(s). (Section 20.2)
extensive property A property that depends on the amount of material considered; for example, mass or volume. (Section 1.3)
functional group An atom or group of atoms that imparts characteristic chemical properties to an organic compound. (Section 24.1)
Hall–Héroult process A process used to obtain aluminum by electrolysis of Al2O3 dissolved in molten cryolite, Na3AlF6. (Section 20.9)
face-centered lattice A crystal lattice in which the lattice points are located at the faces and corners of each unit cell. (Section 12.2)
fusion The joining of two light nuclei to form a more massive one. (Section 21.6)
halogens Members of group 7A in the periodic table. (Section 7.8)
Faraday’s constant (F) The magnitude of charge of one mole of electrons: 96,500 C/mol. (Section 20.5)
gamma radiation Energetic electromagnetic radiation emanating from the nucleus of a radioactive atom. (Section 21.1)
f-block metals Lanthanide and actinide elements in which the 4f or 5f orbitals are partially occupied. (Section 6.9)
gas Matter that has no fixed volume or shape; it conforms to the volume and shape of its container. (Section 1.2)
ferrimagnetism A form of magnetism in which unpaired electron spins on different-type ions point in opposite directions but do not fully cancel out. (Section 23.1)
gas constant (R) The constant of proportionality in the ideal-gas equation. (Section 10.4)
ferromagnetism A form of magnetism in which unpaired electron spins align parallel to one another. (Section 23.1) first law of thermodynamics A statement that energy is conserved in any process. One way to express the law is that the change in internal energy, ¢E, of a system in any process is equal to the heat, q, added to the system, plus the work, w, done on the system by its surroundings: ¢E = q + w. (Section 5.2) first-order reaction A reaction in which the reaction rate is proportional to the concentration of a single reactant, raised to the first power. (Section 14.4) fission The splitting of a large nucleus into two smaller ones. (Section 21.6) folding The process by which a protein adopts its biologically active shape. (Section 24.7) force A push or a pull. (Section 5.1) formal charge The number of valence electrons in an isolated atom minus the number of electrons assigned to the atom in the Lewis structure. (Section 8.5) formation constant For a metal ion complex, the equilibrium constant for formation of the complex from the metal ion and base species present in solution. It is a measure of the tendency of the complex to form. (Section 17.5) formula weight The mass of the collection of atoms represented by a chemical formula. For example, the formula weight of NO2 (46.0 amu) is the sum of the masses of one nitrogen atom and two oxygen atoms. (Section 3.3) fossil fuels Coal, oil, and natural gas, which are presently our major sources of energy. (Section 5.8) free energy (Gibbs free energy, G) A thermodynamic state function that gives a criterion for spontaneous change in terms of enthalpy and entropy: G = H - TS. (Section 19.5) free radical A substance with one or more unpaired electrons. (Section 21.9) frequency The number of times per second that one complete wavelength passes a given point. (Section 6.1) frequency factor (A) A term in the Arrhenius equation that is related to the frequency of collision and the probability that the collisions are favorably oriented for reaction. (Section 14.5) fuel cell A voltaic cell that utilizes the oxidation of a conventional fuel, such as H2 or CH4, in the cell reaction. (Section 20.7)
galvanic cell See voltaic cell. (Section 20.3)
hard water Water that contains appreciable concentrations of Ca2 + and Mg 2 + ; these ions react with soaps to form an insoluble material. (Section 18.4) heat The flow of energy from a body at higher temperature to one at lower temperature when they are placed in thermal contact. (Section 5.1) heat capacity The quantity of heat required to raise the temperature of a sample of matter by 1 °C (or 1 K). (Section 5.5)
geometric isomerism A form of isomerism in which compounds with the same type and number of atoms and the same chemical bonds have different spatial arrangements of these atoms and bonds. (Sections 23.4 and 24.4)
heat of fusion The enthalpy change, ¢H, for melting a solid. (Section 11.4)
Gibbs free energy A thermodynamic state function that combines enthalpy and entropy, in the form G = H - TS. For a change occurring at constant temperature and pressure, the change in free energy is ¢G = ¢H - T¢S. (Section 19.5)
heat of vaporization The enthalpy change, ¢H, for vaporization of a liquid. (Section 11.4)
glass An amorphous solid formed by fusion of SiO2, CaO, and Na2O. Other oxides may also be used to form glasses with differing characteristics. (Section 22.10) glucose A polyhydroxy aldehyde whose formula is CH2OH(CHOH)4CHO; it is the most important of the monosaccharides. (Section 24.8) glycogen The general name given to a group of polysaccharides of glucose that are synthesized in mammals and used to store energy from carbohydrates. (Section 24.7)
heat of sublimation The enthalpy change, ¢H, for vaporization of a solid. (Section 11.4)
Henderson–Hasselbalch equation The relationship among the pH, pKa, and the concentrations of acid and conjugate base in an aqueous [base] solution: pH = pKa + log . (Section 17.2) [acid] Henry’s law A law stating that the concentration of a gas in a solution, Sg, is proportional to the pressure of gas over the solution: Sg = kPg. (Section 13.3) Hess’s law The heat evolved in a given process can be expressed as the sum of the heats of several processes that, when added, yield the process of interest. (Section 5.6)
Graham’s law A law stating that the rate of effusion of a gas is inversely proportional to the square root of its molecular weight. (Section 10.8)
heterogeneous alloy An alloy in which the components are not distributed uniformly; instead, two or more distinct phases with characteristic compositions are present. (Section 12.3)
gray (Gy) The SI unit for radiation dose corresponding to the absorption of 1 J of energy per kilogram of biological material; 1 Gy = 100 rads. (Section 21.9)
heterogeneous catalyst A catalyst that is in a different phase from that of the reactant substances. (Section 14.7)
green chemistry Chemistry that promotes the design and application of chemical products and processes that are compatible with human health and that preserve the environment. (Section 18.5) greenhouse gases Gases in an atmosphere that absorb and emit infrared radiation (radiant heat), “trapping” heat in the atmosphere. (Section 18.2) ground state The lowest-energy, or most stable, state. (Section 6.3) group Elements that are in the same column of the periodic table; elements within the same group or family exhibit similarities in their chemical behavior. (Section 2.5) Haber process The catalyst system and conditions of temperature and pressure developed by Fritz Haber and coworkers for the formation of NH3 from H2 and N2. (Section 15.2) half-life The time required for the concentration of a reactant substance to decrease to half its initial value; the time required for half of a sample of a particular radioisotope to decay. (Sections 14.4 and 21.4) half-reaction An equation for either an oxidation or a reduction that explicitly shows the electrons
heterogeneous equilibrium The equilibrium established between substances in two or more different phases, for example, between a gas and a solid or between a solid and a liquid. (Section 15.4) hexagonal close packing A crystal structure where the atoms are packed together as closely as possible. The close-packed layers adopt a two-layer repeating pattern, which leads to a primitive hexagonal unit cell. (Section 12.3) high-spin complex A complex whose electrons populate the d orbitals to give the maximum number of unpaired electrons. (Section 23.6) hole A vacancy in the valence band of a semiconductor, created by doping. (Section 12.7) homogeneous catalyst A catalyst that is in the same phase as the reactant substances. (Section 14.7) homogeneous equilibrium The equilibrium established between reactant and product substances that are all in the same phase. (Section 15.4) Hund’s rule A rule stating that electrons occupy degenerate orbitals in such a way as to maximize the number of electrons with the same spin. In
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GLOSSARY
other words, each orbital has one electron placed in it before pairing of electrons in orbitals occurs. (Section 6.8) hybridization The mixing of different types of atomic orbitals to produce a set of equivalent hybrid orbitals. (Section 9.5) hybrid orbital An orbital that results from the mixing of different kinds of atomic orbitals on the same atom. For example, an sp3 hybrid results from the mixing, or hybridizing, of one s orbital and three p orbitals. (Section 9.5) hydration Solvation when the solvent is water. (Section 13.1) hydride ion An ion formed by the addition of an electron to a hydrogen atom: H- . (Section 7.7)
compounds that form between metallic elements. (Section 12.3)
joule (J) The SI unit of energy, 1 kg-m2>s2. A related unit is the calorie: 4.184 J = 1 cal. (Section 5.1)
intermolecular forces The short-range attractive forces operating between the particles that make up the units of a liquid or solid substance. These same forces also cause gases to liquefy or solidify at low temperatures and high pressures. (Chapter 11: Introduction)
Kelvin scale The absolute temperature scale; the SI unit for temperature is the kelvin. Zero on the Kelvin scale corresponds to -273.15 °C. (Section 1.4)
internal energy The total energy possessed by a system. When a system undergoes a change, the change in internal energy, ¢E, is defined as the heat, q, added to the system, plus the work, w, done on the system by its surroundings: ¢E = q + w. (Section 5.2)
hydrocarbons Compounds composed of only carbon and hydrogen. (Section 2.9)
interstitial alloy An alloy in which smaller atoms fit into spaces between larger atoms. The larger atoms are metallic elements and the smaller atoms are typically nonmetallic elements. (Section 12.3)
hydrogen bonding Bonding that results from intermolecular attractions between molecules containing hydrogen bonded to an electronegative element. The most important examples involve OH, NH, and HF. (Section 11.2)
ion Electrically charged atom or group of atoms (polyatomic ion); ions can be positively or negatively charged, depending on whether electrons are lost (positive) or gained (negative) by the atoms. (Section 2.7)
hydrolysis A reaction with water. When a cation or anion reacts with water, it changes the pH. (Sections 16.9 and 24.4)
ion–dipole force The force that exists between an ion and a neutral polar molecule that possesses a permanent dipole moment. (Section 11.2)
hydronium ion (H3O+) The predominant form of the proton in aqueous solution. (Section 16.2) hydrophilic Water attracting. The term is often used to describe a type of colloid. (Section 13.6) hydrophobic Water repelling. The term is often used to describe a type of colloid. (Section 13.6) hypothesis A tentative explanation of a series of observations or of a natural law. (Section 1.3) ideal gas A hypothetical gas whose pressure, volume, and temperature behavior is completely described by the ideal-gas equation. (Section 10.4) ideal-gas equation An equation of state for gases that embodies Boyle’s law, Charles’s law, and Avogadro’s hypothesis in the form PV = nRT. (Section 10.4) ideal solution A solution that obeys Raoult’s law. (Section 13.5) immiscible liquids Liquids that do not dissolve in one another to a significant extent. (Section 13.3) indicator A substance added to a solution that changes color when the added solute has reacted with all the solute present in solution. The most common type of indicator is an acid–base indicator whose color changes as a function of pH. (Section 4.6) instantaneous rate The reaction rate at a particular time as opposed to the average rate over an interval of time. (Section 14.2) intensive property A property that is independent of the amount of material considered, for example, density. (Section 1.3)
ion exchange A process by which ions in solution are exchanged for other ions held on the surface of an ion-exchange resin; the exchange of a hard-water cation such as Ca2 + for a soft-water cation such as Na + is used to soften water. (Section 18.4) ionic bond A bond between oppositely charged ions. The ions are formed from atoms by transfer of one or more electrons. (Section 8.1) ionic compound A compound composed of cations and anions. (Section 2.7) ionic hydrides Compounds formed when hydrogen reacts with alkali metals and also the heavier alkaline earths (Ca, Sr, and Ba); these compounds contain the hydride ion, H- . (Section 22.2) ionic solids Solids that are composed of ions. (Section 12.1) ionization energy The energy required to remove an electron from a gaseous atom when the atom is in its ground state. (Section 7.4) ionizing radiation Radiation that has sufficient energy to remove an electron from a molecule, thereby ionizing it. (Section 21.9) ion-product constant For water, Kw is the product of the aquated hydrogen ion and hydroxide ion concentrations: [H + ][OH-] = Kw = 1.0 * 10-14 at 25 °C. (Section 16.3) irreversible process A process that cannot be reversed to restore both the system and its surroundings to their original states. Any spontaneous process is irreversible. (Section 19.1)
interhalogens Compounds formed between two different halogen elements. Examples include IBr and BrF3. (Section 22.4)
isoelectronic series A series of atoms, ions, or molecules having the same number of electrons. (Section 7.3)
intermediate A substance formed in one elementary step of a multistep mechanism and consumed in another; it is neither a reactant nor an ultimate product of the overall reaction. (Section 14.6)
isomers Compounds whose molecules have the same overall composition but different structures. (Sections 2.9 and 23.4)
intermetallic compound A homogeneous alloy with definite properties and a fixed composition. Intermetallic compounds are stoichiometric
isotopes Atoms of the same element containing different numbers of neutrons and therefore having different masses. (Section 2.3)
isothermal process One that occurs at constant temperature. (Section 19.1)
ketone A compound in which the carbonyl group (C “ O) occurs at the interior of a carbon chain and is therefore flanked by carbon atoms. (Section 24.4) kinetic energy The energy that an object possesses by virtue of its motion. (Section 5.1) kinetic-molecular theory A set of assumptions about the nature of gases. These assumptions, when translated into mathematical form, yield the ideal-gas equation. (Section 10.7) lanthanide contraction The gradual decrease in atomic and ionic radii with increasing atomic number among the lanthanide elements, atomic numbers 57 through 70. The decrease arises because of a gradual increase in effective nuclear charge through the lanthanide series. (Section 23.1) lanthanide (rare earth) element Element in which the 4f subshell is only partially occupied. (Sections 6.8 and 6.9) lattice energy The energy required to separate completely the ions in an ionic solid. (Section 8.2) lattice points Points in a crystal all of which have identical environments. (Section 12.2) lattice vectors The vectors a, b, and c that define a crystal lattice. The position of any lattice point in a crystal can be represented by summing integer multiples of the lattice vectors. (Section 12.2) law of constant composition A law that states that the elemental composition of a pure compound is always the same, regardless of its source; also called the law of definite proportions. (Section 1.2) law of definite proportions A law that states that the elemental composition of a pure substance is always the same, regardless of its source; also called the law of constant composition. (Section 1.2) law of mass action The rules by which the equilibrium constant is expressed in terms of the concentrations of reactants and products, in accordance with the balanced chemical equation for the reaction. (Section 15.2) Le Châtelier’s principle A principle stating that when we disturb a system at chemical equilibrium, the relative concentrations of reactants and products shift so as to partially undo the effects of the disturbance. (Section 15.7) levorotatory, or merely levo or l A term used to label a chiral molecule that rotates the plane of polarization of plane-polarized light to the left (counterclockwise). (Section 24.4) Lewis acid An electron-pair acceptor. (Section 16.11) Lewis base An electron-pair donor. (Section 16.11) Lewis structure A representation of covalent bonding in a molecule that is drawn using Lewis symbols. Shared electron pairs are shown as lines, and unshared electron pairs are shown as pairs of dots. Only the valence-shell electrons are shown. (Section 8.3)
GLOSSARY Lewis symbol (electron-dot symbol) The chemical symbol for an element, with a dot for each valence electron. (Section 8.1) ligand An ion or molecule that coordinates to a metal atom or to a metal ion to form a complex. (Section 23.2) lime-soda process A method used in largescale water treatment to reduce water hardness by removing Mg 2 + and Ca2 + . The substances added to the water are lime, CaO [or slaked lime, Ca(OH)2], and soda ash, Na2CO3, in amounts determined by the concentrations of the undesired ions. (Section 18.4) limiting reactant (limiting reagent) The reactant present in the smallest stoichiometric quantity in a mixture of reactants; the amount of product that can form is limited by the complete consumption of the limiting reactant. (Section 3.7) line spectrum A spectrum that contains radiation at only certain specific wavelengths. (Section 6.3)
mean free path The average distance traveled by a gas molecule between collisions. (Section 10.8) metal complex An assembly of a metal ion and the Lewis bases bonded to it. (Section 23.2) metallic bond Bonding, usually in solid metals, in which the bonding electrons are relatively free to move throughout the three-dimensional structure. (Section 8.1) metallic character The extent to which an element exhibits the physical and chemical properties characteristic of metals, for example, luster, malleability, ductility, and good thermal and electrical conductivity. (Section 7.6) metallic elements (metals) Elements that are usually solids at room temperature, exhibit high electrical and heat conductivity, and appear lustrous. Most of the elements in the periodic table are metals. (Sections 2.5 and 12.1)
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molar heat capacity The heat required to raise the temperature of one mole of a substance by 1 °C. (Section 5.5) molarity The concentration of a solution expressed as moles of solute per liter of solution; abbreviated M. (Section 4.5) molar mass The mass of one mole of a substance in grams; it is numerically equal to the formula weight in atomic mass units. (Section 3.4) mole A collection of Avogadro’s number (6.022 * 1023) of objects; for example, a mole of H2O is 6.022 * 1023 H2O molecules. (Section 3.4) molecular compound A compound that consists of molecules. (Section 2.6) molecular equation A chemical equation in which the formula for each substance is written without regard for whether it is an electrolyte or a nonelectrolyte. (Section 4.2)
metallic hydrides Compounds formed when hydrogen reacts with transition metals; these compounds contain the hydride ion, H- . (Section 22.2)
molecular formula A chemical formula that indicates the actual number of atoms of each element in one molecule of a substance. (Section 2.6)
metallic solids Solids that are composed of metal atoms. (Section 12.1)
molecular geometry The arrangement in space of the atoms of a molecule. (Section 9.2) molecular hydrides Compounds formed when hydrogen reacts with nonmetals and metalloids. (Section 22.2)
liquid Matter that has a distinct volume but no specific shape. (Section 1.2)
metalloids Elements that lie along the diagonal line separating the metals from the nonmetals in the periodic table; the properties of metalloids are intermediate between those of metals and nonmetals. (Section 2.5)
liquid crystal A substance that exhibits one or more partially ordered liquid phases above the melting point of the solid form. By contrast, in nonliquid crystalline substances the liquid phase that forms upon melting is completely unordered. (Section 11.7)
metallurgy The science of extracting metals from their natural sources by a combination of chemical and physical processes. It is also concerned with the properties and structures of metals and alloys. (Section 23.1)
lock-and-key model A model of enzyme action in which the substrate molecule is pictured as fitting rather specifically into the active site on the enzyme. It is assumed that in being bound to the active site, the substrate is somehow activated for reaction. (Section 14.7)
metathesis (exchange) reaction A reaction in which two substances react through an exchange of their component ions: AX + BY ¡ AY + BX. Precipitation and acid–base neutralization reactions are examples of metathesis reactions. (Section 4.2)
low-spin complex A metal complex in which the electrons are paired in lower-energy orbitals. (Section 23.6)
metric system A system of measurement used in science and in most countries. The meter and the gram are examples of metric units. (Section 1.4)
linkage isomers Structural isomers of coordination compounds in which a ligand differs in its mode of attachment to a metal ion. (Section 23.4) lipid A nonpolar molecule derived from glycerol and fatty acids that is used by organisms for longterm energy storage. (Section 24.9)
magic numbers Numbers of protons and neutrons that result in very stable nuclei. (Section 21.2) main-group elements Elements in the s and p blocks of the periodic table. (Section 6.9) mass A measure of the amount of material in an object. It measures the resistance of an object to being moved. In SI units, mass is measured in kilograms. (Section 1.4) mass defect The difference between the mass of a nucleus and the total masses of the individual nucleons that it contains. (Section 21.6) mass number The sum of the number of protons and neutrons in the nucleus of a particular atom. (Section 2.3) mass percentage The number of grams of solute in each 100 g of solution. (Section 13.4) mass spectrometer An instrument used to measure the precise masses and relative amounts of atomic and molecular ions. (Section 2.4)
microstate The state of a system at a particular instant; one of many possible energetically equivalent ways to arrange the components of a system to achieve a particular state. (Section 19.3)
molecularity The number of molecules that participate as reactants in an elementary reaction. (Section 14.6) molecular orbital (MO) An allowed state for an electron in a molecule. According to molecularorbital theory, a molecular orbital is entirely analogous to an atomic orbital, which is an allowed state for an electron in an atom. Most bonding molecular orbitals can be classified as s or p, depending on the disposition of electron density with respect to the internuclear axis. (Section 9.7) molecular-orbital diagram A diagram that shows the energies of molecular orbitals relative to the atomic orbitals from which they are derived; also called an energy-level diagram. (Section 9.7) molecular-orbital theory A theory that accounts for the allowed states for electrons in molecules. (Section 9.7) molecular solids Solids that are composed of molecules. (Sections 12.1 and 12.6)
mineral A solid, inorganic substance occurring in nature, such as calcium carbonate, which occurs as calcite. (Section 23.1)
molecular weight The mass of the collection of atoms represented by the chemical formula for a molecule. (Section 3.3)
miscible liquids Liquids that mix in all proportions. (Section 13.3)
molecule A chemical combination of two or more atoms. (Sections 1.1 and 2.6)
mixture A combination of two or more substances in which each substance retains its own chemical identity. (Section 1.2)
mole fraction The ratio of the number of moles of one component of a mixture to the total moles of all components; abbreviated X, with a subscript to identify the component. (Section 10.6)
molal boiling-point-elevation constant (Kb) A constant characteristic of a particular solvent that gives the increase in boiling point as a function of solution molality: ¢Tb = Kbm. (Section 13.5)
matter Anything that occupies space and has mass; the physical material of the universe. (Section 1.1)
molal freezing-point-depression constant (Kf) A constant characteristic of a particular solvent that gives the decrease in freezing point as a function of solution molality: ¢Tf = Kf m. (Section 13.5)
matter waves The term used to describe the wave characteristics of a moving particle. (Section 6.4)
molality The concentration of a solution expressed as moles of solute per kilogram of solvent; abbreviated m. (Section 13.4)
momentum The product of the mass, m, and velocity, v, of an object. (Section 6.4) monodentate ligand A ligand that binds to the metal ion via a single donor atom. It occupies one position in the coordination sphere. (Section 23.3) monomers Molecules with low molecular weights, which can be joined together (polymerized) to form a polymer. (Section 12.8) monosaccharide A simple sugar, most commonly containing six carbon atoms. The joining together of monosaccharide units by condensation
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GLOSSARY
reactions results in formation of polysaccharides. (Section 24.8) nanomaterial A solid whose dimensions range from 1 to 100 nm and whose properties differ from those of a bulk material with the same composition. (Section 12.1) natural gas A naturally occurring mixture of gaseous hydrocarbon compounds composed of hydrogen and carbon. (Section 5.8) nematic liquid crystalline phase A liquid crystal in which the molecules are aligned in the same general direction, along their long axes, but in which the ends of the molecules are not aligned. (Section 11.7) Nernst equation An equation that relates the cell emf, E, to the standard emf, E°, and the reaction quotient, Q : E = E° - (RT>nF) ln Q. (Section 20.6) net ionic equation A chemical equation for a solution reaction in which soluble strong electrolytes are written as ions and spectator ions are omitted. (Section 4.2) neutralization reaction A reaction in which an acid and a base react in stoichiometrically equivalent amounts; the neutralization reaction between an acid and a metal hydroxide produces water and a salt. (Section 4.3) neutron An electrically neutral particle found in the nucleus of an atom; it has approximately the same mass as a proton. (Section 2.3) noble gases Members of group 8A in the periodic table. (Section 7.8) node Points in an atom at which the electron density is zero. For example, the node in a 2s orbital is a spherical surface. (Section 6.6) nonbonding pair In a Lewis structure a pair of electrons assigned completely to one atom; also called a lone pair. (Section 9.2) nonelectrolyte A substance that does not ionize in water and consequently gives a nonconducting solution. (Section 4.1) nonionizing radiation Radiation that does not have sufficient energy to remove an electron from a molecule. (Section 21.9) nonmetallic elements (nonmetals) Elements in the upper right corner of the periodic table; nonmetals differ from metals in their physical and chemical properties. (Section 2.5) nonpolar covalent bond A covalent bond in which the electrons are shared equally. (Section 8.4) normal boiling point The boiling point at 1 atm pressure. (Section 11.5) normal melting point The melting point at 1 atm pressure. (Section 11.6) nuclear binding energy The energy required to decompose an atomic nucleus into its component protons and neutrons. (Section 21.6) nuclear disintegration series A series of nuclear reactions that begins with an unstable nucleus and terminates with a stable one; also called a radioactive series. (Section 21.2) nuclear model Model of the atom with a nucleus containing protons and neutrons and with electrons in the space outside the nucleus. (Section 2.2) nuclear transmutation A conversion of one kind of nucleus to another. (Section 21.3)
nucleic acids Polymers of high molecular weight that carry genetic information and control protein synthesis. (Section 24.10)
oxyacid A compound in which one or more OH groups, and possibly additional oxygen atoms, are bonded to a central atom. (Section 16.10)
nucleon A particle found in the nucleus of an atom. (Section 21.1)
oxyanion A polyatomic anion that contains one or more oxygen atoms. (Section 2.8)
nucleotide Compounds formed from a molecule of phosphoric acid, a sugar molecule, and an organic nitrogen base. Nucleotides form linear polymers called DNA and RNA, which are involved in protein synthesis and cell reproduction. (Section 24.10)
ozone The name given to O3, an allotrope of oxygen. (Section 7.8)
nucleus The very small, very dense, positively charged portion of an atom; it is composed of protons and neutrons. (Section 2.2)
paramagnetism A property that a substance possesses if it contains one or more unpaired electrons. A paramagnetic substance is drawn into a magnetic field. (Section 9.8) partial pressure The pressure exerted by a particular gas in a mixture. (Section 10.6)
octet rule A rule stating that bonded atoms tend to possess or share a total of eight valence-shell electrons. (Section 8.1)
particle accelerator A device that uses strong magnetic and electrostatic fields to accelerate charged particles. (Section 21.3)
optical isomerism A form of isomerism in which the two forms of a compound (stereoisomers) are nonsuperimposable mirror images. (Section 23.4)
parts per billion (ppb) The concentration of a solution in grams of solute per 109 (billion) grams of solution; equals micrograms of solute per liter of solution for aqueous solutions. (Section 13.4)
optically active Possessing the ability to rotate the plane of polarized light. (Section 23.4)
parts per million (ppm) The concentration of a solution in grams of solute per 106 (million) grams of solution; equals milligrams of solute per liter of solution for aqueous solutions. (Section 13.4)
orbital An allowed energy state of an electron in the quantum mechanical model of the atom; the term orbital is also used to describe the spatial distribution of the electron. An orbital is defined by the values of three quantum numbers: n, l, and ml (Section 6.5) organic chemistry The study of carboncontaining compounds, typically containing carbon–carbon bonds. (Section 2.9; Chapter 24: Introduction) osmosis The net movement of solvent through a semipermeable membrane toward the solution with greater solute concentration. (Section 13.5) osmotic pressure The pressure that must be applied to a solution to stop osmosis from pure solvent into the solution. (Section 13.5) Ostwald process An industrial process used to make nitric acid from ammonia. The NH3 is catalytically oxidized by O2 to form NO; NO in air is oxidized to NO2; HNO3 is formed in a disproportionation reaction when NO2 dissolves in water. (Section 22.7) overall reaction order The sum of the reaction orders of all the reactants appearing in the rate expression when the rate can be expressed as rate = k3A4a3B4b. . . . (Section 14.3) overlap The extent to which atomic orbitals on different atoms share the same region of space. When the overlap between two orbitals is large, a strong bond may be formed. (Section 9.4)
pascal (Pa) The SI unit of pressure: 1 Pa = 1 N>m2. (Section 10.2) Pauli exclusion principle A rule stating that no two electrons in an atom may have the same four quantum numbers (n, l, ml, and ms)). As a reflection of this principle, there can be no more than two electrons in any one atomic orbital. (Section 6.7) peptide bond A bond formed between two amino acids. (Section 24.7) percent ionization The percent of a substance that undergoes ionization on dissolution in water. The term applies to solutions of weak acids and bases. (Section 16.6) percent yield The ratio of the actual (experimental) yield of a product to its theoretical (calculated) yield, multiplied by 100. (Section 3.7) period The row of elements that lie in a horizontal row in the periodic table. (Section 2.5) periodic table The arrangement of elements in order of increasing atomic number, with elements having similar properties placed in vertical columns. (Section 2.5) petroleum A naturally occurring combustible liquid composed of hundreds of hydrocarbons and other organic compounds. (Section 5.8) pH The negative log in base 10 of the aquated hydrogen ion concentration: pH = -log3H + 4. (Section 16.4)
oxidation A process in which a substance loses one or more electrons. (Section 4.4)
pH titration curve A graph of pH as a function of added titrant. (Section 17.3)
oxidation number (oxidation state) A positive or negative whole number assigned to an element in a molecule or ion on the basis of a set of formal rules; to some degree it reflects the positive or negative character of that atom. (Section 4.4)
phase change The conversion of a substance from one state of matter to another. The phase changes we consider are melting and freezing (solid Δ liquid), sublimation and deposition, and vaporization and condensation (liquid Δ gas). (Section 11.4)
oxidation-reduction (redox) reaction A chemical reaction in which the oxidation states of certain atoms change. (Section 4.4; Chapter 20: Introduction) oxidizing agent, or oxidant The substance that is reduced and thereby causes the oxidation of some other substance in an oxidation-reduction reaction. (Section 20.1)
phase diagram A graphic representation of the equilibria among the solid, liquid, and gaseous phases of a substance as a function of temperature and pressure. (Section 11.6) phospholipid A form of lipid molecule that contains charged phosphate groups. (Section 24.9)
GLOSSARY photochemical smog A complex mixture of undesirable substances produced by the action of sunlight on an urban atmosphere polluted with automobile emissions. The major starting ingredients are nitrogen oxides and organic substances, notably olefins and aldehydes. (Section 18.2) photodissociation The breaking of a molecule into two or more neutral fragments as a result of absorption of light. (Section 18.2)
polysaccharide A substance made up of many monosaccharide units joined together. (Section 24.8) porphyrin A complex derived from the porphine molecule. (Section 23.3) positron emission A nuclear decay process where a positron, a particle with the same mass as an electron but with a positive charge, symbol 01e, is emitted from the nucleus. (Section 21.1)
photoelectric effect The emission of electrons from a metal surface induced by light. (Section 6.2)
potential energy The energy that an object possesses as a result of its composition or its position with respect to another object. (Section 5.1)
photoionization The removal of an electron from an atom or molecule by absorption of light. (Section 18.2)
precipitate An insoluble substance that forms in, and separates from, a solution. (Section 4.2)
photon The smallest increment (a quantum) of radiant energy; a photon of light with frequency n has an energy equal to hn. (Section 6.2) photosynthesis The process that occurs in plant leaves by which light energy is used to convert carbon dioxide and water to carbohydrates and oxygen. (Section 23.3) physical changes Changes (such as a phase change) that occur with no change in chemical composition. (Section 1.3) physical properties Properties that can be measured without changing the composition of a substance, for example, color and freezing point. (Section 1.3) pi (p) bond A covalent bond in which electron density is concentrated above and below the internuclear axis. (Section 9.6) pi (p) molecular orbital A molecular orbital that concentrates the electron density on opposite sides of an imaginary line that passes through the nuclei. (Section 9.8) Planck’s constant (h) The constant that relates the energy and frequency of a photon, E = hn. Its value is 6.626 * 10-34 J-s. (Section 6.2) plastic A material that can be formed into particular shapes by application of heat and pressure. (Section 12.8) polar covalent bond A covalent bond in which the electrons are not shared equally. (Section 8.4) polarizability The ease with which the electron cloud of an atom or a molecule is distorted by an outside influence, thereby inducing a dipole moment. (Section 11.2) polar molecule A molecule that possesses a nonzero dipole moment. (Section 8.4) polyatomic ion An electrically charged group of two or more atoms. (Section 2.7) polydentate ligand A ligand in which two or more donor atoms can coordinate to the same metal ion. (Section 23.3) polymer A large molecule of high molecular mass, formed by the joining together, or polymerization, of a large number of molecules of low molecular mass. The individual molecules forming the polymer are called monomers. (Sections 12.1 and 12.8) polypeptide A polymer of amino acids that has a molecular weight of less than 10,000. (Section 24.7) polyprotic acid A substance capable of dissociating more than one proton in water; H2SO4 is an example. (Section 16.6)
precipitation reaction A reaction that occurs between substances in solution in which one of the products is insoluble. (Section 4.2) precision The closeness of agreement among several measurements of the same quantity; the reproducibility of a measurement. (Section 1.5) pressure A measure of the force exerted on a unit area. In chemistry, pressure is often expressed in units of atmospheres (atm) or torr: 760 torr = 1 atm; in SI units pressure is expressed in pascals (Pa). (Section 10.2) pressure–volume (PV) work Work performed by expansion of a gas against a resisting pressure. (Section 5.3) primary structure The sequence of amino acids along a protein chain. (Section 24.7) primitive lattice A crystal lattice in which the lattice points are located only at the corners of each unit cell. (Section 12.2) probability density (C 2) A value that represents the probability that an electron will be found at a given point in space. Also called electron density. (Section 6.5) product A substance produced in a chemical reaction; it appears to the right of the arrow in a chemical equation. (Section 3.1) property A characteristic that gives a sample of matter its unique identity. (Section 1.1) protein A biopolymer formed from amino acids. (Section 24.7) protium The most common isotope of hydrogen. (Section 22.2) proton A positively charged subatomic particle found in the nucleus of an atom. (Section 2.3) pure substance Matter that has a fixed composition and distinct properties. (Section 1.2) pyrometallurgy A process in which heat converts a mineral in an ore from one chemical form to another and eventually to the free metal. (Section 23.2) qualitative analysis The determination of the presence or absence of a particular substance in a mixture. (Section 17.7) quantitative analysis The determination of the amount of a given substance that is present in a sample. (Section 17.7) quantum The smallest increment of radiant energy that may be absorbed or emitted; the magnitude of radiant energy is hn. (Section 6.2) quaternary structure The structure of a protein resulting from the clustering of several individual protein chains into a final specific shape. (Section 24.7)
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racemic mixture A mixture of equal amounts of the dextrorotatory and levorotatory forms of a chiral molecule. A racemic mixture will not rotate the plane of polarized light. (Section 23.4) rad A measure of the energy absorbed from radiation by tissue or other biological material; 1 rad = transfer of 1 * 10 - 2 J of energy per kilogram of material. (Section 21.9) radial probability function The probability that the electron will be found at a certain distance from the nucleus. (Section 6.6) radioactive Possessing radioactivity, the spontaneous disintegration of an unstable atomic nucleus with accompanying emission of radiation. (Section 2.2; Chapter 21: Introduction) radioactive series A series of nuclear reactions that begins with an unstable nucleus and terminates with a stable one. Also called nuclear disintegration series. (Section 21.2) radioisotope An isotope that is radioactive; that is, it is undergoing nuclear changes with emission of radiation. (Section 21.1) radionuclide A radioactive nuclide. (Section 21.1) radiotracer A radioisotope that can be used to trace the path of an element in a chemical system. (Section 21.5) Raoult’s law A law stating that the partial pressure of a solvent over a solution, Psolution, is given by ° the vapor pressure of the pure solvent, P solvent , times the mole fraction of a solvent in the solution, ° . (Section 13.5) Xsolvent: Psolution = XsolventP solvent rare earth element See lanthanide element. (Sections 6.8 and 6.9) rate constant A constant of proportionality between the reaction rate and the concentrations of reactants that appear in the rate law. (Section 14.3) rate-determining step The slowest elementary step in a reaction mechanism. (Section 14.6) rate law An equation that relates the reaction rate to the concentrations of reactants (and sometimes of products also). (Section 14.3) reactant A starting substance in a chemical reaction; it appears to the left of the arrow in a chemical equation. (Section 3.1) reaction mechanism A detailed picture, or model, of how the reaction occurs; that is, the order in which bonds are broken and formed and the changes in relative positions of the atoms as the reaction proceeds. (Section 14.6) reaction order The power to which the concentration of a reactant is raised in a rate law. (Section 14.3) reaction quotient (Q) The value that is obtained when concentrations of reactants and products are inserted into the equilibrium expression. If the concentrations are equilibrium concentrations, Q = K; otherwise, Q Z K. (Section 15.6) reaction rate A measure of the decrease in concentration of a reactant or the increase in concentration of a product with time. (Section 14.2) redox (oxidation-reduction) reaction A reaction in which certain atoms undergo changes in oxidation states. The substance increasing in oxidation state is oxidized; the substance decreasing in oxidation state is reduced. (Section 4.4; Chapter 20: Introduction) reducing agent, or reductant The substance that is oxidized and thereby causes the reduction of
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GLOSSARY
some other substance in an oxidation-reduction reaction. (Section 20.1) reduction A process in which a substance gains one or more electrons. (Section 4.4) rem A measure of the biological damage caused by radiation; rems = rads * RBE. (Section 21.9) renewable energy sources Energy such as solar energy, wind energy, and hydroelectric energy derived from essentially inexhaustible sources. (Section 5.8) representative (main-group) element An element from within the s and p blocks of the periodic table (Figure 6.29). (Section 6.9) resonance structures (resonance forms) Individual Lewis structures in cases where two or more Lewis structures are equally good descriptions of a single molecule. The resonance structures in such an instance are “averaged” to give a more accurate description of the real molecule. (Section 8.6) reverse osmosis The process by which water molecules move under high pressure through a semipermeable membrane from the more concentrated to the less concentrated solution. (Section 18.4) reversible process A process that can go back and forth between states along exactly the same path; a system at equilibrium is reversible if equilibrium can be shifted by an infinitesimal modification of a variable such as temperature. (Section 19.1) ribonucleic acid (RNA) A polynucleotide in which ribose is the sugar component. (Section 24.10) root-mean-square (rms) speed (M) The square root of the average of the squared speeds of the gas molecules in a gas sample. (Section 10.7) rotational motion Movement of a molecule as though it is spinning like a top. (Section 19.3) salinity A measure of the salt content of seawater, brine, or brackish water. It is equal to the mass in grams of dissolved salts present in 1 kg of seawater. (Section 18.3) salt An ionic compound formed by replacing one or more hydrogens of an acid by other cations. (Section 4.3) saponification Hydrolysis of an ester in the presence of a base. (Section 24.4) saturated solution A solution in which undissolved solute and dissolved solute are in equilibrium. (Section 13.2)
second-order reaction A reaction in which the overall reaction order (the sum of the concentration-term exponents) in the rate law is 2. (Section 14.4) sigma (S) bond A covalent bond in which electron density is concentrated along the internuclear axis. (Section 9.6) sigma (S) molecular orbital A molecular orbital that centers the electron density about an imaginary line passing through two nuclei. (Section 9.7) significant figures The digits that indicate the precision with which a measurement is made; all digits of a measured quantity are significant, including the last digit, which is uncertain. (Section 1.5) silica Common name for silicon dioxide. (Section 22.4) silicates Compounds containing silicon and oxygen, structurally based on SiO4 tetrahedra. (Section 22.10) single bond A covalent bond involving one electron pair. (Section 8.3) SI units The preferred metric units for use in science. (Section 1.4) smectic liquid crystalline phase A liquid crystal in which the molecules are aligned along their long axes and arranged in sheets, with the ends of the molecules aligned. There are several different kinds of smectic phases. (Section 12.8) solid Matter that has both a definite shape and a definite volume. (Section 1.2) solubility The amount of a substance that dissolves in a given quantity of solvent at a given temperature to form a saturated solution. (Sections 4.2 and 13.2) solubility-product constant (solubility product) (Ksp) An equilibrium constant related to the equilibrium between a solid salt and its ions in solution. It provides a quantitative measure of the solubility of a slightly soluble salt. (Section 17.4) solute A substance dissolved in a solvent to form a solution; it is normally the component of a solution present in the smaller amount. (Section 4.1) solution A mixture of substances that has a uniform composition; a homogeneous mixture. (Section 1.2)
scientific law A concise verbal statement or a mathematical equation that summarizes a wide range of observations and experiences. (Section 1.3)
solution alloy A homogeneous alloy, where two or more elements are distributed randomly and uniformly throughout the solid. (Section 12.3)
scientific method The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws. (Section 1.3)
solvation The clustering of solvent molecules around a solute particle. (Section 13.1)
secondary structure The manner in which a protein is coiled or stretched. (Section 24.7) second law of thermodynamics A statement of our experience that there is a direction to the way events occur in nature. When a process occurs spontaneously in one direction, it is nonspontaneous in the reverse direction. It is possible to state the second law in many different forms, but they all relate back to the same idea about spontaneity. One of the most common statements found in chemical contexts is that in any spontaneous process the entropy of the universe increases. (Section 19.2)
solvent The dissolving medium of a solution; it is normally the component of a solution present in the greater amount. (Section 4.1) specific heat (Cs) The heat capacity of 1 g of a substance; the heat required to raise the temperature of 1 g of a substance by 1 °C. (Section 5.5) spectator ions Ions that go through a reaction unchanged and that appear on both sides of the complete ionic equation. (Section 4.2) spectrochemical series A list of ligands arranged in order of their abilities to split the d-orbital energies (using the terminology of the crystalfield model). (Section 23.6)
spectrum The distribution among various wavelengths of the radiant energy emitted or absorbed by an object. (Section 6.3) spin magnetic quantum number (ms) A quantum number associated with the electron spin; it may have values of + 12 or - 12 . (Section 6.7) spin-pairing energy The energy required to pair an electron with another electron occupying an orbital. (Section 23.6) spontaneous process A process that is capable of proceeding in a given direction, as written or described, without needing to be driven by an outside source of energy. A process may be spontaneous even though it is very slow. (Section 19.1) standard atmospheric pressure Defined as 760 torr or, in SI units, 101.325 kPa. (Section 10.2) standard emf, also called the standard cell potential (E°) The emf of a cell when all reagents are at standard conditions. (Section 20.4) standard enthalpy change (¢H°) The change in enthalpy in a process when all reactants and products are in their stable forms at 1 atm pressure and a specified temperature, commonly 25 °C. (Section 5.7) standard enthalpy of formation (¢Hf°) The change in enthalpy that accompanies the formation of one mole of a substance from its elements, with all substances in their standard states. (Section 5.7) standard free energy of formation (¢Gf°) The change in free energy associated with the formation of a substance from its elements under standard conditions. (Section 19.5) standard hydrogen electrode (SHE) An electrode based on the half-reaction 2 H + (1 M) + 2 e - ¡ H2(1 atm). The standard electrode potential of the standard hydrogen electrode is defined as 0 V. (Section 20.4) standard molar entropy (S°) The entropy value for a mole of a substance in its standard state. (Section 19.4) ° ) The postandard reduction potential (Ered tential of a reduction half-reaction under standard conditions, measured relative to the standard hydrogen electrode. A standard reduction potential is also called a standard electrode potential. (Section 20.4)
standard solution A solution of known concentration. (Section 4.6) standard temperature and pressure (STP) Defined as 0 °C and 1 atm pressure; frequently used as reference conditions for a gas. (Section 10.4) starch The general name given to a group of polysaccharides that acts as energy-storage substances in plants. (Section 24.8) state function A property of a system that is determined by its state or condition and not by how it got to that state; its value is fixed when temperature, pressure, composition, and physical form are specified; P, V, T, E, and H are state functions. (Section 5.2) states of matter The three forms that matter can assume: solid, liquid, and gas. (Section 1.2) stereoisomers Compounds possessing the same formula and bonding arrangement but differing in the spatial arrangements of the atoms. (Section 23.4) stoichiometry The relationships among the quantities of reactants and products involved in chemical reactions. (Chapter 3: Introduction)
GLOSSARY stratosphere The region of the atmosphere directly above the troposphere. (Section 18.1) strong acid An acid that ionizes completely in water. (Section 4.3) strong base A base that ionizes completely in water. (Section 4.3) strong electrolyte A substance (strong acids, strong bases, and most salts) that is completely ionized in solution. (Section 4.1) structural formula A formula that shows not only the number and kinds of atoms in the molecule but also the arrangement (connections) of the atoms. (Section 2.6) structural isomers Compounds possessing the same formula but differing in the bonding arrangements of the atoms. (Sections 23.4 and 24.2) subatomic particles Particles such as protons, neutrons, and electrons that are smaller than an atom. (Section 2.2) subshell One or more orbitals with the same set of quantum numbers n and l. For example, we speak of the 2p subshell (n = 2, l = 1), which is composed of three orbitals (2px, 2py, and 2pz). (Section 6.5) substitutional alloy A homogeneous (solution) alloy in which atoms of different elements randomly occupy sites in the lattice. (Section 23.6) substitution reactions Reactions in which one atom (or group of atoms) replaces another atom (or group) within a molecule; substitution reactions are typical for alkanes and aromatic hydrocarbons. (Section 24.3) substrate A substance that undergoes a reaction at the active site in an enzyme. (Section 14.7) supercritical mass An amount of fissionable material larger than the critical mass. (Section 21.7) supersaturated solution A solution containing more solute than an equivalent saturated solution. (Section 13.2) surface tension The intermolecular, cohesive attraction that causes a liquid to minimize its surface area. (Section 11.3) surroundings In thermodynamics, everything that lies outside the system that we study. (Section 5.1) system In thermodynamics, the portion of the universe that we single out for study. We must be careful to state exactly what the system contains and what transfers of energy it may have with its surroundings. (Section 5.1) termolecular reaction An elementary reaction that involves three molecules. Termolecular reactions are rare. (Section 14.6) tertiary structure The overall shape of a large protein, specifically, the manner in which sections of the protein fold back upon themselves or intertwine. (Section 24.7) theoretical yield The quantity of product that is calculated to form when all of the limiting reagent reacts. (Section 3.7) theory A tested model or explanation that satisfactorily accounts for a certain set of phenomena. (Section 1.3)
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thermochemistry The relationship between chemical reactions and energy changes. (Chapter 5: Introduction)
valence band A band of closely spaced molecular orbitals that is essentially fully occupied by electrons. (Section 12.7)
thermodynamics The study of energy and its transformation. (Chapter 5: Introduction)
valence-bond theory A model of chemical bonding in which an electron-pair bond is formed between two atoms by the overlap of orbitals on the two atoms. (Section 9.4)
thermonuclear reaction Another name for fusion reactions; reactions in which two light nuclei are joined to form a more massive one. (Section 21.8) thermoplastic A polymeric material that can be readily reshaped by application of heat and pressure. (Section 12.8) thermosetting plastic A plastic that, once formed in a particular mold, is not readily reshaped by application of heat and pressure. (Section 12.8) third law of thermodynamics A law stating that the entropy of a pure, crystalline solid at absolute zero temperature is zero: S(0 K) = 0. (Section 19.3) titration The process of reacting a solution of unknown concentration with one of known concentration (a standard solution). (Section 4.6) torr A unit of pressure (1 torr = 1 mm Hg). (Section 10.2) transition elements (transition metals) Elements in which the d orbitals are partially occupied. (Section 6.8) transition state (activated complex) The particular arrangement of reactant and product molecules at the point of maximum energy in the rate-determining step of a reaction. (Section 14.5) translational motion Movement in which an entire molecule moves in a definite direction. (Section 19.3) transuranium elements Elements that follow uranium in the periodic table. (Section 21.3) triple bond A covalent bond involving three electron pairs. (Section 8.3) triple point The temperature at which solid, liquid, and gas phases coexist in equilibrium. (Section 11.6) tritium The isotope of hydrogen whose nucleus contains a proton and two neutrons. (Section 22.2) troposphere The region of Earth’s atmosphere extending from the surface to about 12 km altitude. (Section 18.1) Tyndall effect The scattering of a beam of visible light by the particles in a colloidal dispersion. (Section 13.6) uncertainty principle A principle stating there is an inherent uncertainty in the precision with which we can simultaneously specify the position and momentum of a particle. This uncertainty is significant only for particles of extremely small mass, such as electrons. (Section 6.4)
valence electrons The outermost electrons of an atom; those that occupy orbitals not occupied in the nearest noble-gas element of lower atomic number. The valence electrons are the ones the atom uses in bonding. (Section 6.8) valence orbitals Orbitals that contain the outershell electrons of an atom. (Chapter 7: Introduction) valence-shell electron-pair repulsion (VSEPR) model A model that accounts for the geometric arrangements of shared and unshared electron pairs around a central atom in terms of the repulsions between electron pairs. (Section 9.2) van der Waals equation An equation of state for nonideal gases that is based on adding corrections to the ideal-gas equation. The correction terms account for intermolecular forces of attraction and for the volumes occupied by the gas molecules themselves. (Section 10.9) vapor Gaseous state of any substance that normally exists as a liquid or solid. (Section 10.1) vapor pressure The pressure exerted by a vapor in equilibrium with its liquid or solid phase. (Section 11.5) vibrational motion Movement of the atoms within a molecule in which they move periodically toward and away from one another. (Section 19.3) viscosity A measure of the resistance of fluids to flow. (Section 11.3) volatile Tending to evaporate readily. (Section 11.5) voltaic (galvanic) cell A device in which a spontaneous oxidation-reduction reaction occurs with the passage of electrons through an external circuit. (Section 20.3) vulcanization The process of cross-linking polymer chains in rubber. (Section 12.6) watt A unit of power; 1 W = 1 J>s. (Section 20.5) wave function A mathematical description of an allowed energy state (an orbital) for an electron in the quantum mechanical model of the atom; it is usually symbolized by the Greek letter c. (Section 6.5) wavelength The distance between identical points on successive waves. (Section 6.1)
unimolecular reaction An elementary reaction that involves a single molecule. (Section 14.6)
weak acid An acid that only partly ionizes in water. (Section 4.3)
unit cell The smallest portion of a crystal that reproduces the structure of the entire crystal when repeated in different directions in space. It is the repeating unit or building block of the crystal lattice. (Section 12.2)
weak base A base that only partly ionizes in water. (Section 4.3)
unsaturated solution A solution containing less solute than a saturated solution. (Section 13.2)
work The movement of an object against some force. (Section 5.1)
weak electrolyte A substance that only partly ionizes in solution. (Section 4.1)
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PHOTO AND ART CREDITS CHAPTER 1: CO01 J. Hester/NASA 1.2a iStockphoto 1.2b Francis G. Mayer/Corbis 1.2c Shutterstock 1.3 Eric Schrader - Pearson Science 1.7 Charles D. Winters/Photo Researchers, Inc. 1.8a Sergej Petrakov/iStockphoto 1.8b Richard Megna/Fundamental Photographs 1.11a Pearson Education 1.11b-c Richard Megna/Fundamental Photographs 1.12a-b Richard Megna/Fundamental Photographs 1.14a-c Richard Megna/Fundamental Photographs 1.15 Eric Schrader - Pearson Science 1.20 AP Photo/Paul Sancya 1.22 Dwight Cendrowski/Cree, Inc. p. 23 Sidney Harris p. 34 (top) sciencephotos/Alamy p. 34 (middle) Shutterstock CHAPTER 2: CO02 Charles D. Winters/Photo Researchers, Inc. 2.1 Corbis/Bettmann 2.2 Wilson Ho/University of California Irvine 2.3a-b Richard Megna/Fundamental Photographs 2.6 Radium Institute/Emilio Segre Archives 2.7 Reserve Bank of New Zealand 2.16a-b Richard Megna/Fundamental Photographs 2.17 University of California Berkeley/Emilio Segre Archives 2.21 iStockphoto 2.23 Eric Schrader - Pearson Science CHAPTER 3: CO03 Shutterstock 3.1 Jean-Loup Charmet/SPL/Photo Researchers, Inc. 3.4 Shutterstock 3.6a-c Richard Megna/Fundamental Photographs 3.7 Donald Johnston/Getty Images 3.8 Richard Megna/Fundamental Photographs 3.10 Richard Megna/ Fundamental Photographs 3.11 iStockphoto p. 109 AP Photo/Damian Dovarganes p. 110 Richard Megna/Fundamental Photographs p. 111 iStockphoto CHAPTER 4: CO04 Macduff Everton/Getty Images 4.2a-c Eric Schrader - Pearson Science 4.4a-c Richard Megna/Fundamental Photographs 4.5 Eric Schrader - Pearson Education 4.8 Eric Schrader - Pearson Science 4.9a-c Richard Megna/Fundamental Photographs 4.10 Eric Schrader - Pearson Science 4.11a-c iStockphoto 4.12a-b Richard Megna/ Fundamental Photographs 4.13a-c Eric Schrader - Pearson Science 4.14a-c Richard Megna/Fundamental Photographs 4.15 AP Photo/Jim Cole 4.16a-c Pearson Education 4.17a-c Richard Megna/Fundamental Photographs 4.19a-d Richard Megna/Fundamental Photographs p. 154 Richard Megna/Fundamental Photographs CHAPTER 5: CO05 Paulo Fridman/Bloomberg via Getty Images 5.1a REUTERS/Adam Hunger 5.1b iStockphoto 5.8a-b Richard Megna/Fundamental Photographs 5.14a-b Charles D. Winters/Photo Researchers, Inc. 5.15 Corbis/Bettmann 5.20 iStockphoto 5.24 Eric Schrader - Pearson Science 5.26 iStockphoto p. 196 James Urbach/Photolibrary CHAPTER 6: CO06 Gary Corbett/AGE Photostock 6.1 Pal Hermansen/Getty Images 6.5 Peter Arnold/AGE Fotostock 6.8 Emilio Segre Archives 6.10a-b Richard Megna/ Fundamental Photographs 6.13 Dr. Nicola Pinna 6.14 Emilio Segre Archives 6.28 Medical Body Scans/Photo Researchers, Inc. p. 240 (top left) iStockphoto p. 240 (bottom) Getty Images p. 240 (right) iStockphoto CHAPTER 7: CO07 Steven Puetzer/Getty Images 7.13 iStockphoto 7.15b Richard Megna/Fundamental Photographs 7.16 SPL/Photo Researchers, Inc. 7.17a-b Richard Megna/Fundamental Photographs 7.18 Andrew Lambert Photography/Photo Researchers, Inc. 7.19 Charles D. Winters/Photo Researchers, Inc. 7.20a-c Richard Megna/ Fundamental Photographs 7.21a-b David Taylor/Photo Researchers, Inc. 7.21c Andrew Lambert Photography/Photo Researchers, Inc. 7.22 iStockphoto 7.23 Write Stuff Syndicate 7.24 Richard Megna/Fundamental Photographs 7.25 Eric Schrader - Pearson Science 7.27 Richard Megna/Fundamental Photographs CHAPTER 8: CO08 Barry O’Neill 8.1 Shutterstock 8.16 Bettmann/Corbis CHAPTER 9: CO09 Richard Megna/Fundamental Photographs 9.5a-c Kristen Brochmann/Fundamental Photographs 9.29 Science Photo Library RF/Photolibrary 9.45 Richard Megna/Fundamental Photographs 9.47 Michael Graetzel
CHAPTER 12: CO12 iStockphoto 12.2a iStockphoto 12.2b Dorling Kindersley Media Library 12.10 Eric Schrader - Pearson Science 12.16 DoITPoMS Micrograph Library, University of Cambridge 12.19a Katharine “GoldenRuby” 12.19b Anne Hänsel 12.25 iStockphoto 12.34 Grace Harbor Industries 12.36 Richard Megna/Fundamental Photographs 12.43 Horst Weller 12.44 Horst Weller 12.45 Photo by Vassil. (666vassil@ orange.fr) 12.46 The Royal Institution of Great Britain/Bridgeman p. 504 (middle) iStockphoto p. 504 (bottom) Jinghong Li CHAPTER 13: CO13 Clark Little Photography 13.5 Tom Bochsler - Pearson Education 13.6a-c Richard Megna/Fundamental Photographs 13.7 Richard Megna/Fundamental Photographs 13.9a-c Richard Megna/Fundamental Photographs 13.10 Richard Megna/ Fundamental Photographs 13.16 Charles D. Winters/Photo Researchers, Inc. 13.17 iStockphoto 13.22 iStockphoto 13.28 Richard Megna/Fundamental Photographs 13.29a iStockphoto 13.29b Shutterstock 13.33 Oliver Meckes & Nicole Ottawa/Photo Researchers, Inc. CHAPTER 14: CO14 Philippe Plailly/SPL/Photo Researchers, Inc. 14.1a Michael S. Yamashita/Corbis 14.1b-c Shutterstock 14.2a Michael Dalton/Fundamental Photographs 14.2b Richard Megna/Fundamental Photographs 14.13 Richard Megna/Fundamental Photographs 14.22a-c Richard Megna/Fundamental Photographs 14.26 Richard Megna/ Fundamental Photographs 14.30 Science Photo Library/Photo Researchers, Inc. CHAPTER 15: CO15 Bruno Perousse/AGE Fotostock 15.1a-c Richard Megna/Fundamental Photographs 15.4 iStockphoto p. 631 Public Domain 15.13a-c Richard Megna/Fundamental Photographs CHAPTER 16: CO16 Charles Mann 16.6 Richard Megna/Fundamental Photographs 16.8a-c Richard Megna/Fundamental Photographs 16.10a-d Eric Schrader - Pearson Science 16.14 Eric Schrader - Pearson Science 16.15 Richard Megna/Fundamental Photographs p. 694 Eric Schrader - Pearson Science CHAPTER 17: CO17 Corbis/Photolibrary 17.1 EMD Chemicals 17.5 P. Motta and S. Corer/SPL/Photo Researchers, Inc. 17.18 Pacific Stock/Photolibrary 17.20a-c Eric Schrader - Pearson Science 17.21a-c Richard Megna/Fundamental Photographs 17.22a-c Richard Megna/Fundamental Photographs p. 740 Richard Megna/Fundamental Photographs CHAPTER 18: CO18 Shutterstock 18.2 iStockphoto 18.5 Karin Jackson/U.S. Air Force/USGS 18.6 NASA 18.7 National Atmospheric Deposition Program 18.8a Hulton Archive/Getty Images 18.8b Dan Breckwoldt/iStockphoto 18.10 iStockphoto 18.14 Astralia Picture Library/Corbis 18.18 iStockphoto 18.19 E. I. Dupont de Nemours & Co. 18.21 Vestergaard Frandsen 18.22 Sheila Terry/SPL/Photo Researchers, Inc. 18.23 Los Angeles Department of Water and Power CHAPTER 19: CO19 iStockphoto 19.1a-b Kristen Brochmann/Fundamental Photographs 19.3a-b Michael Dalton/Fundamental Photographs 19.7 Osterreichische Zentralbibliothek fur Physik 19.12a GM Pictures/Alamy 19.12b Ben Levy - RapidRepair.com 19.15 Manuscripts & Archives - Yale University Library 19.18 iStockphoto CHAPTER 20: CO20 Shutterstock 20.1a-b Richard Megna/Fundamental Photographs 20.2a-b Richard Megna/Fundamental Photographs 20.3a-b Richard Megna/Fundamental Photographs 20.4 Richard Megna/Fundamental Photographs 20.7 iStockphoto 20.13 Michael Faraday, oil, by Thomas Phillips 20.28 Oberlin College Archives
CHAPTER 10: CO10 NASA 10.4 Pearson Education 10.12 Andrew Putler/Redferns/ Getty Images 10.20b Richard Megna/Fundamental Photographs
CHAPTER 21: CO21 JPL/NASA 21.5 Brookhaven National Laboratory 21.8 Don Murray/Getty Images 21.11 Drs. Suzanne Baker, William Jagust, and Susan Landau 21.17 Los Alamos National Laboratory 21.20 Idaho National Laboratory 21.26 Environmental Protection Agency
CHAPTER 11: CO11 Tongro Image Stock/AGE Fotostock 11.1 Bharat Bhushan/Ohio State University 11.2a Leslie Garland Picture Library/Alamy 11.2b Dorling Kindersley 11.2c Dennis “S.K” used under a Creative Commons license. http://creativecommons.org/ licenses/by/3.0/ p. 433 Bill Waterson/Universal Press Syndicate 11.11 Ted Kinsman/ Photo Researchers, Inc. 11.12 Eric Schrader - Pearson Science 11.17 Kristen Brochmann/ Fundamental Photographs 11.18 Hermann Eisenbeiss/Photo Researchers, Inc. 11.19 Richard Megna/Fundamental Photographs 11.31a-b Richard Megna/Fundamental Photographs 11.35 REUTERS/Yuriko Nakao p. 458 Richard Megna/Fundamental Photographs
CHAPTER 22: CO22 Feng Yu/Fotolia 22.5a-b Richard Megna/Fundamental Photographs 22.8 Pearson Education 22.10a-b Richard Megna/Fundamental Photographs 22.11 NASA/Johnson Space Center 22.12 Prof. Carl Djerassi 22.13 Shutterstock 22.14a-b Richard Megna/Fundamental Photographs 22.15 Maksym Gorpenyuk/Fotolia 22.17 Corbis RF/Alamy 22.19 Paul Silverman/Fundamental Photographs 22.20a-b Kristen Brochmann/Fundamental Photographs 22.24a-b Richard Megna/Fundamental Photographs 22.25a-b Pearson Education 22.27 Richard Megna/Fundamental Photographs 22.30 United States Geological Service 22.31 Shutterstock 22.32 Eric Schrader - Pearson Science 22.35 National Institute for Occupational Safety and Health
P-1
P-2
CREDITS
CHAPTER 23: CO23 photo and restoration by Jamie Rife - Rife Glass 23.3 Richard Megna/Fundamental Photographs 23.6 Joel Arem/Photo Researchers, Inc. 23.7 Richard Megna/Fundamental Photographs 23.8a-b Richard Megna/Fundamental Photographs 23.20a-b Richard Megna/Fundamental Photographs 23.24 Richard Megna/Fundamental Photographs 23.26 Dr. Nigel J. Forrow 23.35a-c Eric Schrader - Pearson Science p. 997 Shutterstock p. 1000 (left) Eric Schrader - Pearson Science p. 1000 (right) Eric Schrader Pearson Science
CHAPTER 24: CO24 Shutterstock 24.6 Shutterstock 24.12 Richard Megna/Fundamental Photographs 24.13a-b Eric Schrader - Pearson Science 24.19 Eric Schrader - Pearson Science
INDEX ABn molecules, 332–33 Absolute entropy, 800, 801, 802 Absolute temperature, 403, 578 Absolute zero, 390 Absorption, 542, 590 Absorption spectrum, 987 Accident prevention, safer chemistry for, 772 Accuracy, 21–22 Acetaldehyde (ethanal), 554, 1022, 1024, 1025 properties of, 431 Acetaminophen, 1049 Acetate ion, 63 Acetic acid (ethanoic acid), 124, 667, 688, 700, 704, 1022, 1025–26 chemical formula of, 118n decarbonylation of, 822 in green chemistry, 773 hydrogen bonding in, 435 ionization of, 118 pH of, 670–71 production of, 824, 1026 properties of, 667 pure (glacial acetic acid), 153 shape of, 342 titration with caustic soda (NaOH), 716–20 Acetic anhydride, 1048 Acetone (propanone), 150, 457, 460, 520, 554, 555, 1022, 1024, 1025 Acetonitrile, 430, 551, 553, 573, 600, 1006 methyl isonitrate converted to, 571, 577–78, 600 properties of, 431 Acetylene, 203–4, 329, 822, 949, 1008, 1017, 1022 production of, 419 standard enthalpy of formation for, 184 triple bonds in, 353–54 Acetylide ion, 948 Acetylsalicylic acid (aspirin), 4, 111, 127n, 377, 697, 747, 1025 Acid(s), 124–25 adipic, 102–3, 494 Arrhenius, 652 binary, 685–86 carboxylic, 688–89 conjugate, 655 defined, 64 defined constants for, 1062 diprotic, 124, 664 as electron acceptor, 689–90 factors affecting strength of, 685 household, 124 ionic properties of, 124–25 Lewis, 689–92 metal oxides reacting with, 266 monoprotic, 124, 664 names and formulas of, 64–65 oxidation of metals by, 133–35 oxyacids, 686–88, 929 properties of, 124–25 reactions of. See also Acid-base reactions with alcohol, 646
with magnesium, 134 with magnesium hydroxide, 128 relating to anions, 64 relative strengths of, 656–58 strong, 125–26, 656–57, 664–66 in buffered solutions, 711–13 titrating, 145–46 weak, 125–26, 656–57, 666–75 acid-dissociation constant (Ka), 667–68, 670–73, 679–81 common-ion effect on, 704–7 percent ionization of, 669 polyprotic, 674–75, 720–21 Acid-base equilibria, 650–701. See also Aqueous equilibria acid-dissociation constant and basedissociation constant relationship, 679–81 Arrhenius definitions, 652 autoionization of water, 658–60 Brønsted-Lowry acids and bases, 652–58 conjugate acid-base pairs, 654–56 H+ ions in water and, 652–53 proton-transfer reactions, 653–54 relative strengths of, 656–58 chemical structure and, 685–89 binary acids, 685–86 carboxylic acids, 688–89 factors affecting acid strength, 685 oxyacids, 686–88 ion product, 659–60 Lewis acids and bases, 689–92 electron-pair acceptor/donor concept, 689–90 metal ions and, 690 in organic substances, 689 pH scale, 660–64 measuring, 663–64 other “p” scales, 662–63 of salt solutions, 681–85 anion reaction with water, 681–82 cation reaction with water, 682–83 combined cation-anion effect, 683–85 solubility equilibria and, 726 strong acids and bases, 664–66 in buffered solutions, 711–13 weak acids and bases, 666–79 acid-dissociation constant, 667–68, 670–73, 679–81 common-ion effect on, 704–7 percent ionization of, 669 polyprotic acids, 674–75 types of, 677–79 Acid-base indicators, 145–46, 663–64 Acid-base properties, of organic substances, 1007 Acid-base reactions, 124–30. See also Acidbase equilibria electrolytes, 126–27 with gas formation, 129–30 gas-phase, 654 neutralization reactions and salts, 127–29 Acid-base titrations, 714–22 of polyprotic acids, 720–21
strong, 714–16 weak, 716–20 Acid-dissociation constant (Ka), 667–68 base-dissociation constant (Kb) and, 679–81 calculating from pH, 668–69 calculating percent ionization using, 673 calculating pH from, 670–73 for metal cations, 682 for polyprotic acids, 674 Acidic anhydrides (acidic oxides), 932 Acidic solutions, 683–85 Acid inhibitors, 130 Acid-insoluble sulfides, 738 Acidosis, 713 Acid rain, 268, 758–60, 932 Acid salts, 680 Acid spills, 130 Actinides, electron configurations of, 232–33 Activated complex (transition state), 577, 582 Activation, entropy of, 825 Activation energy, 577–78 catalysis and, 590, 593, 637, 638 determining, 579–81 Active metals, 135 Active site, 593 Active transport, 539 Activity, 888–89 Activity series, 135–37, 846 Actual yield, 102–3 Addition in exponential notation, 1052 significant figures in, 24 Addition polymerization, 492–93 Addition polymers, 493 Addition reactions of alkenes and alkynes, 1017–19 mechanism of, 1019 Adenine, 459, 1040, 1041 Adenosine diphosphate (ADP), 814, 944 Adenosine monophosphate (AMP), 1050 Adenosine triphosphate (ATP), 814, 824, 944 Adhesive forces, 438 Adipic acid, 102–3, 494 ADP (adenosine diphosphate), 814, 944 Adrenaline (epinephrine), 108, 552, 698 Adsorption, 542, 543, 590–91 Advanced burning, 903 Air, 113 combustion in, 83–84 composition of, 384, 400 density of, 19 Air bags, automobile, 83, 398–99 Air pollution, 592 in Mexico City, 778 Alanine, 689, 1030, 1031, 1032 Alanylglycylserine, 1032 (R)-Albuterol, 1029 Alchemy, 138 Alcohol(s), 66–67, 1023–24 condensation reactions with, 1026 as functional groups, 1023 oxidation of, 1025
reaction of organic acid with, 646 solubilities of, 521 Aldehydes, 1024–25 nomenclature, 1047 Alizarin yellow R, 664 Alkali chlorides, lattice energies of, 294 Alkali metal halides, properties of, 481 Alkali metal ions, 738 Alkali (group 1A) metals, 51, 135, 233, 268–72 group trends for, 268–72 in ionic hydrides, 923 ionic hydroxides of, 665–66 oxidation number of, 132 as reducing agent, 844 Alkaline battery, 855–56 Alkaline earth (group 2A) metals, 51, 135, 233, 272 group trends for, 272 in ionic hydrides, 923 ionic hydroxides of, 665–66 oxidation number of, 132 as reducing agent, 844 Alkalosis, 713 Alkanes, 66–67, 1008, 1009–14 cycloalkanes, 1013 derivatives of, 66–67 nomenclature of, 1010–13 reactions of, 1013–14 structural isomers of, 1009–10 structures of, 1009 Alka-Seltzer, 130 Alkenes, 1008, 1015–17 addition reactions of, 1017–19 Alkyl groups, 1011, 1021 in Friedel-Crafts reaction, 1021 Alkynes, 1008, 1017–19 addition reactions of, 1017–19 in green chemistry, 774–75 Allene, 379 Allotropes, 273 Alloys, 473–76, 964 common, 473 defined, 473 heterogeneous, 474, 475 intermetallic compounds, 475 interstitial, 474, 475 solution, 474 steels, 474 substitutional, 474, 475, 476 Alloy steels, 474 a-helix, 1033 a radiation, 43–44, 877, 878, 879, 902, 903, 904 a-scattering experiment, Rutherford’s, 43–44 Aluminum (Al), 8, 249, 476, 506 alloyed with gold, 477 electrometallurgy of, 862 electron configuration of, 233 Lewis symbol for, 290 oxidation number of, 132 oxidation of, 136, 857–58 purification of ore, 733–34 recycling, 862 specific heat of, 176 thermodynamic quantities for, 1059
I-1
I-2
Index
Aluminum hydroxide, 734 Aluminum ion (Al3+), 60 Aluminum oxide, 862 American Chemical Society, 52 Americium–241 606 Amides, 1028 Amine group, 689 Amine hydrochlorides, 680 Amines, 677–78, 680, 1007, 1028 reaction with carboxylic acid, 493–94 Amino acids, 1029–32 a–, 1030 amphiprotic behavior of, 689 chiral, 1030 essential, 1030 general structure of, 689 side chain, 1032 Ammeter, 873 Ammonia, 124, 384, 442, 738, 937–38 as Arrhenius and Brønsted-Lowry base, 654 bond angles, 338 bonding in, 350 critical temperature and pressure of, 442 in fertilizers, 614, 615 formation of, 803 Haber (Haber-Bosch) process for synthesizing, 614, 615, 873 free energy changes in, 810, 813 hydrogen and, 923 nitrogen and, 937 temperature effects on, 615, 631 as household base, 124 molecular geometry of, 336 properties of, 676, 678 reactions of with boron trifluoride, 313 with water, 655 sale of, 6 standard enthalpy of formation for, 184 synthesis of, 614, 615, 630–31, 637–38, 802, 813 Ammonium chloride, 422 Ammonium cyanate, 1006 Ammonium hydroxide, 125 Ammonium ions, 60, 119 reaction with nitrite ions in water, 563 Ammonium nitrate, 516, 517 Ammonium perchlorate, 929–30, 960 Ammonium thiocyanate, 168 Amorphous solid, 465 AMP (adenosine monophosphate), 1050 Ampere (A or amp), 15 Amphetamine hydrochloride, 680 Amphiprotic substances, 654, 733n Amphojel, 130 Amphoteric hydroxides, 733–34 Amphoteric oxides, 733–34 Amphoteric substances, 932 Amphoterism, 726, 733–34 amu (atomic mass unit), 45, 48 Anaerobic reaction, 192 Anemia iron-deficiency, 978 sickle-cell, 545 -ane suffix, 66 Angina pectoris, 193, 941 Angstrom (Å), 45, 208
Angular momentum quantum number, 220 Anhydrides acidic, 932 basic, 932 Aniline (phenylamine), 1028 Anion(s), 54 borane, 953 carboxylate, 688 chemical formulas and, 119 combined effect with cation, 683–85 common, 61, 62, 63 in freshwater, 767 in ionic liquids, 436 names and formulas of, 61–62 oxygen, 274 reaction with water, 681–82 relating to acids, 64 size of, 256–57 Anode, 258, 836–37 sacrificial, 859 Antacids, 130, 155 Anthracene, 381, 1019 Anthracite coal, 190 Antibiotics, bacterial resistance to, modification to combat, 21 Antibonding molecular orbital, 358–59, 360n Anticancer drugs, 379 Antiferromagnetism, 967–68 Antifreeze, 530, 535 Antiknock agents, 1014 Antilogarithms, 1054 Antimony, 941 Aqua regia, 138 Aqueous equilibria, 702–47 acid-base titrations, 714–22 of polyprotic acids, 720–21 strong, 714–16 weak, 716–20 buffered solutions, 707–13 blood as, 713 buffer capacity and pH, 710–11 calculating pH of buffer, 708–10 composition and action of, 707–8 strong acids or bases in, 711–13 common-ion effect, 704–7 solubility and, 726–27 precipitation and separation of ions, 734–36 in qualitative analysis for metallic elements, 736–39 solubility equilibria, 722–26 solubility-product constant (Ksp), 722–26 Aqueous reactions. See also Solution(s) of ions, 734–36 molarity, 527–30 oxidation-reduction, 131–38 activity series and, 135–37 defined, 131 molecular and net ionic equations for, 135 oxidation numbers (oxidation states), 132–33 oxidation of metals by acids and salts, 133–35 precipitation, 119–24 exchange (metathesis) reactions, 121–22 ionic equations, 122–24
solubility guidelines for ionic compounds, 120–21 selective, 735–36 stoichiometry acid-base reactions, 124–30 electrolytes, 126–27 with gas formation, 129–30 neutralization reactions and salts, 127–29 Aqueous solution(s), 114–57, 513. See also Acid-base equilibria activity series of metals in, 135–37 defined, 115 electrolysis of, 860 electrolytic properties of, 116–17 freezing-point depression in, 535 ionic compounds in water, 117–18 molecular compounds in water, 118 strong and weak electrolytes, 118–19 of transition metal ions, 966 Aquifers, 767 Arginine, 1031 Argon (Ar), 49, 232, 442, 924 in air, 384 in atmosphere, 751 critical temperature and pressure of, 442 electron affinity of, 263–64 Lewis symbol for, 290 properties of, 276 Aristotle, 40 Aromatic hydrocarbons, 1008, 1019–21 Aromatic molecules, 311 Array detector, 468 Arrhenius, Svante, 577, 652 Arrhenius acids and bases, 652 Arrhenius equation, 578–79 catalysis and, 590 Arsenic, 941, 961 in drinking water, 156, 526, 945 in groundwater, 767 Arsenic(III) sulfide, 420 Asbestos, 951 serpentine, 951 Ascorbic acid (vitamin C), 127n, 522, 551, 674, 1007, 1025 acid-dissociation constant of, 674 Asparagine, 1031 Aspartame, 108, 1031–32 Aspartic acid, 1031, 1032 Asphalt, 1014 Aspirin (acetylsalicylic acid), 4, 111, 127n, 377, 697, 747, 1025 Astatine, 51, 274, 283 isotopes of, 926 -ate suffix, 61, 62 Atmosphere, 382, 750–64 carbon dioxide in, 696, 701, 703–4 composition of, 750–52 methyl bromide in, 574, 757 nitrogen oxides and photochemical smog, 592, 760–61 ozone in, 754–56 depletion of, 756–57 regions of, 750 sulfur compounds and acid rain, 758–60 temperature of, 750 troposphere, 750 water vapor, carbon dioxide and climate, 761–64
Atmosphere (atm), 386 Atmospheric pressure, 169n, 172, 385–87 standard, 386 Atmospheric residence time, 574 Atom(s), 4 Dalton’s definition of, 40 donor, 974 of element, 8 elements and, 40 estimating number of, 87 many-electron, 226–29 nuclear model of, 43–44 sizes of, 254–59 Atom economy, 771, 774–75 Atomic bomb, 408, 897–98, 902n Atomic emission, 213 Atomic Energy Commission, 52 Atomic masses, 48. See also Stoichiometry Atomic mass scale, 47–48 Atomic mass unit (amu), 45, 48 Atomic number, 46–47, 50, 251, 876 Atomic orbitals, 219–22 crystal-field theory and, 988–89 d, 225–26 energies of, 226 f, 225–26 interactions between 2s and 2p, 365–66 p, 224–25, 252 phases in, 363–64 quantum numbers and, 220–22 representations of, 222–26 s, 222–24 wave functions, 364 Atomic perspective, 4–5 Atomic radius/radii, 254–59 bonding (covalent), 254 bond lengths and, 254–55 nonbonding (van der Waals), 254 periodic trends in, 255–56 Atomic structure. See also Electronic structure discovery of, 41–44 cathode rays and electrons, 41–43 nuclear atom, 43–44 radioactivity, 43 modern view of, 44–47 atomic numbers, mass numbers, isotopes, 46–47 “plum pudding” model of, 43 Atomic theory of matter, 40–41 Atomic weights, 47–49, 85, 251 Atomium, 288 Atomization, 315 Atomos, 40 ATP (adenosine triphosphate), 814, 824, 944 Atto prefix, 16 Aurora borealis, 751 Autoionization of water, 658–60 ion product, 659–60 Automobiles. See Cars Average atomic masses, 48 Averages, calculating, 1057 Average speed, 403 Avogadro, Amedeo, 87, 390 Avogadro’s hypothesis, 390 Avogadro’s law, 390–91, 394 Avogadro’s number, 86–92 Axial electron domains, 339 Azide ion, 379 Azides, in green chemistry, 774–75
Index Azobenzene, 380 Azo dyes, 380 Backbone, carbon-carbon, 1007 Background radiation, 905 Bacteria antibiotic-resistant, 21 iron metabolism in, 978–79 Badge dosimeter, 891 Baeyer-Villiger reaction, 780 Baking soda, 124, 947 Balanced equations, 78–80 for combination and decomposition reactions, 83 quantitative information from, 96–99 Ball-and-stick model, 54, 332 Balloons hot-air, 390 weather, 388 “Ballpark” estimate, 26 Balmer, Johann, 213 Balsa wood, density of, 19 Band gap, 487, 488–89 Bands of energy states, of metals, 479 Band structure, 480 Bangladesh, arsenic in water in, 945 Bar, 385, 801n Barium (Ba), 272 electron configuration of, 232, 244 oxidation in aqueous solution, 136 properties of, 272 thermodynamic quantities for, 1059 Barium azide, 329 Barium chloride, mole relationships of, 88 Barium chloride dihydrate, 518 Barium hydroxide, 932 Barium hydroxide octahydrate, 168 Barium ion (Ba2+), 60 Barium oxide, 932 Barometer, 385–87 Bartlett, Neil, 276 Base(s), 124, 125–26. See also Acid-base reactions amphoteric, 733–34 Arrhenius, 652 conjugate, 655 defined, 125 dissociation constants for, 1063 household, 124 Lewis, 689–92 relative strengths of, 656–58 strong, 125–26, 665–66 in buffered solutions, 711–13 weak, 125–26, 676–79 common-ion effect of, 704–7 types of, 677–79 Base-dissociation constant (Kb), 676 acid-dissociation constant (Ka) and, 679–81 Base-insoluble sulfides and hydroxides, 738 Base units, 15 Basic anhydrides (basic oxides), 932 Basic solutions, 683–85 balancing equations for reactions in, 833–35 Batteries, 826, 827, 854–57 alkaline, 855–56 defined, 854 fuel cells, 856–57 lead-acid, 855 lithium ion, 258
nickel-cadmium, nickel-metal hydride, and lithium-ion, 856 oxidation-reduction in, 855 primary and secondary, 855 in series, 854 sodium ion, 284 Battery acid, 855 Bauxite, 734, 862 Be2, 361–62 Becquerel (Bq), 888 Becquerel, Henri, 43, 891 Beer’s law, 564 Bellamy, Hillary, 143 Belousov-Zhabotinsky reaction, 556 Belt of stability, 880–81 “Bends,” the (decompression sickness), 525 Bent geometry, 333, 334, 337, 343 Benzamide, 1028 Benzene, 72, 203–4, 532, 1008, 1019 bonding in, 1020 bromination of, 1021 hydrogenation of, 1020 isomers of, 1021 molal boiling-point elevation and freezing-point depression constants of, 534 properties of, 486 resonance in, 311–12 sigma and pi bond networks in, 355–56 standard enthalpy of formation for, 184 in styrene manufacture, 772 Benzenesulfonic acid, 746 Benzocaine, 1025 Benzoic acid (phenyl methanoic acid), 179, 667, 688, 697, 1025 properties of, 667 Bernoulli, Daniel, 405 Bertholet, Marcellin, 788 Beryllium (Be), 903 effective nuclear charge of, 253 electron affinity of, 264 electron configuration of, 230, 233 ionization energy of, 261 Lewis symbol for, 290 melting and boiling points of, 428 properties of, 272 thermodynamic quantities for, 1059 Berzelius, Jons Jakob, 490 b radiation, 43, 878, 879, 902, 904 Beta sheet, 1033 BHT (butylated hydroxytoluene), 553 Bicarbonate ion, 713 Bidentate ligand, 974 Big Bang, 903 Big Island of Hawaii, 748 Bile, 544 Bimolecular reaction, 581, 583 Binary acids, 685–86 Binary hydrides, 686 Binary hydrogen compounds, 923–24 Binary molecular compounds, names and formulas of, 65–66 Biochemistry, 1006, 1029 carbohydrates, 1034–37 disaccharides, 1035–36 monosaccharides, 1035–36 polysaccharides, 1036–37 entropy and, 1029 lipids, 1037–38
nucleic acids, 1038–42 proteins. See Protein(s) Biodegradable material, 768 Biodiesel, 192 Bioenergy, 158 Bioethanol, 192 Biofuels, 20 scientific and political challenges of, 192 Biological chemistry. See Biochemistry Biological effects of radiation, 900, 902, 904 dosage and, 904–5 radon, 906 therapeutic, 875, 893, 907 Biological fluids, pH of, 661 Biological systems, phosphorus compounds in, 944. See also Biochemistry; Living systems Biomass energy, 20, 191 Biopolymers, 1029 Bioremediation, 646, 782 Bipolar affective disorder (manicdepressive illness), 271 bi-prefix, 62 Bismuth, 235, 276–77 properties of, 941 Bismuth subsalicylate, 276 Bisulfates (hydrogen sulfates), 936 Bisulfites, 935 Bituminous coal, 190 Blackbody radiation, 210 Blood as buffered solution, 707, 713 metal ion complexes in, 1002 pH range of, 661 Blood alcohol concentration, 153 Blood gases, deep-sea diving and, 525 Blood pressure, 388 Blood sugar, 89, 188. See also Glucose monitoring of, 90 Body-centered cubic unit cell, 467, 469, 470 Body temperature, 180 Bohr, Niels, 213–16, 217, 246 Bohr’s model, 213–16 energy states of hydrogen atom, 214–16 limitations of, 216 Boiling point(s) of halogens, 429 intermolecular forces and, 428 molecular weight and, 431 of noble gases, 429 normal, 444, 811–12 vapor pressure and, 444 Boiling-point elevation, 533–34 Boiling water reactor, 899 Boltzmann, Ludwig, 793, 795 Boltzmann’s equation, 794–96 Bomb (constant volume) calorimetry, 178–79 Bond(s) and bonding, 288–329. See also Molecular orbitals (MO) carbon-carbon, 491, 494–95, 948, 950, 1007, 1009, 1015, 1037. See also under Carbon-carbon bonds carbon-hydrogen, 1006, 1007 covalent, 289, 290, 296–98, 304 bond enthalpies and strength of, 316–18 intermolecular forces vs., 427
I-3
octet rule exceptions, 312–14 orbital overlap and, 345–46 polar vs. nonpolar, 299 strengths of, 315–21 dipole moments, 301–3 double, 298 length of, 318–20 in retinal, 357 rotation about, 1016 in double bonds, 352 8–N, 476 electronegativity and, 298–304 hydrogen, 431–34 in water, 432, 433–34 ionic, 289, 290, 291–96, 300, 304, 481 electron configurations of ions and, 294–96 energetics of formation, 292–94 polyatomic ions, 306–7 transition-metal ions, 296 Lewis structures, 297–98 drawing, 305–9 formal charge, 307–9 Lewis symbols, 290 metallic, 289, 290, 468–69, 476–81 electron-sea model for, 478 molecular-orbital model for (band theory), 478–81 metal–ligand, 971–72, 987 multiple, 298 bond angles and, 338–39 electron domains for, 338–39 Lewis structure with, 306 molecular geometry and, 351–58 octet rule, 290–91 exceptions to, 312–14 in oxygen, 930 peptide, 1030–32 pi (p), 352–58 in chemistry of vision, 357 delocalized, 356 polarity, 298–304 resonance structures, 309–12 in benzene, 311–12 in nitrate ion, 310 in ozone, 309–10 sigma (s), 351–52, 354 single, 298 length of, 318–20 rotations about, 1009 in transitional-metal complexes. See Crystal-field theory triple, 298, 353–54 hybrid orbitals and, 353–54 length of, 318–20 in triple bonds, 352 valence electrons and, 312–14 Bond angles, 332 ideal, 336 nonbonding electrons and multiple bonds and, 338–39 predicting, 342–43 Bond dipole, 343 Bond energy, 753–54 Bond enthalpies, 315–21 average, 317–18 bond length and, 318–21 enthalpies of reactions and, 316–18 Bonding atomic radius (covalent radius), 254 Bonding electrons, 359
I-4
Index
Bonding molecular orbital, 358 Bonding pair, 334, 338–39 Bond length(s), 298 atomic radii and, 254–55 bond enthalpy and, 318–21 electronegativity and, 303 Bond order, 360 Bond polarity, 298–304, 343–45 electronegativity and, 300–301 Borane anions, 953 Boranes, 953 Borax, 953 Borazine, 961 Boric acid, 237 in seawater, 765 Boric oxide, 953 Born, Max, 295 Born-Haber cycle, 295 Boron (B), 51, 261, 903, 953–54 electron configuration of, 230, 233 isotopes of, 237 Lewis symbol for, 290 Boron hydrides, 203 Boron nitride, 486 Boron trifluoride, 312–13 Bosch, Karl, 615 Boyle, Robert, 388 Boyle’s law, 388–89, 391, 394, 404, 405 Bragg, William and Lawrence, 510 Branched-chain hydrocarbons, 1009 Brass, 473, 551 Brazil, ethanol production in, 158, 159, 192 British thermal unit (btu), 197, 205 Brittleness, in ionic crystals, 481 Bromate ion, 771 Bromcresol green, 746 Bromide, in seawater, 765 Bromide ion, 63, 589–90 Bromine (Br), 8, 291, 926, 927 in atmosphere, 757 properties of, 274, 275 reaction of with nitric oxide, 586–88 state at room temperature and standard pressure, 427 thermodynamic quantities for, 1059 Bromobenzene, 1021 2-Bromopentane enantiomers, 1028 Bromthymol blue, 664 Brønsted, Johannes, 652 Brønsted acids and bases, 652–58 conjugate acid-base pairs, 654–56 H+ ions in water and, 652–53 proton-transfer reactions, 653–54 relative strengths of, 656–58 Bronze, 473 btu (British thermal unit), 197, 205 Buckminster Fuller, R., 499 Buckminsterfullerene (buckyball), 499 Buffer capacity, 710–11 Buffered solutions, 707–13 blood as, 707, 713 buffer capacity and pH, 710–11 calculating pH of buffer, 708–10 composition and action of, 707–8 strong acids or bases in, 711–13 Burets, 18, 19 Burning, advanced, 903 Butadiene, 379 Butane, 461, 1008, 1009, 1010 combustion of, 97–98, 762
Lewis structure and condensed structural formula for, 1009 in natural gas, 190 Butanoic acid (butyric acid), 1027 Butanol, solubility of, 521 2-Butanone (methyl ethyl ketone), 1024 1-Butene, 1015 2-Butene, 381, 1019 cis-2-Butene, 1015 trans-2-Butene, 1015 Butylated hydroxytoluene (BHT), 553 Butyl chloride, 561 tert-Butyl group, 1011 Butyl group, 1011 1-Butyl-3-methylimidazolium cation, 436 2-Butyne, 1018 Butyric acid, 700, 1027 Cadaverine, 108, 680 Cade, John, 271 Cadmium ion, 60 Cadmium phosphide, 497 Cadmium telluride, 508, 509 Cadmium yellow, 993 CaF2, 723, 725, 726, 727 Caffeine, 108, 551, 680 Calcite, 726, 948 Calcium (Ca), 8, 506 electron configuration of, 233 as essential nutrient, 272 oxidation of, 131, 132, 136 properties of, 272 reaction with water, 272 in seawater, 765 thermodynamic quantities for, 1059 Calcium carbide, 948–49 Calcium carbonate (limestone), 6, 272, 393, 728 corrosion of by acid rain, 758–59 decomposition of, 83, 624 dissolution of, 728 exoskeleton made of, 703, 728 precipitation of, 722 reaction with sulfur dioxide, 759 specific heat of, 176 standard enthalpy of formation for, 184 Calcium chloride, 535 Calcium hydride, 419, 923 Calcium ions, 58, 60 Calcium oxide (lime or quicklime), 83, 759, 948 standard enthalpy of formation for, 184 Calcium sulfite, 759 Calculations involving many variables, 393 significant figures in, 23–25 Calculators, 451 Caloric content, 189 Calorie (Cal), 162 calorie (cal), 162 Calorimeter, 175 coffee-cup, 177–78 Calorimetry, 175–79 bomb (constant-volume), 178–79 constant-pressure, 177–78 heat capacity and specific heat, 175–77 Cancer lung, 906 from radiation, 904, 906
radiation therapy for, 875, 907 Cancer agents, in water, 771 Candela (cd), 15 Capillary action, 438 Caproic acid, 699 Caraway, 1025 Carbides, 948–49 Carbohydrates, 188, 189–90, 1034–37 as biofuels, 192 disaccharides, 1035–36 monosaccharides, 1035–36 polysaccharides, 1036–37 Carbon (C), 8, 58, 945–49. See also Carbon dioxide; Carbon monoxide bonds about, 1009 carbides, 948–49 carbonic acid and carbonates, 948 electron configuration of, 230 elemental forms of, 945–46 formation of, 903 fullerenes, 498–99 graphene, 499–501 inorganic compounds of, 949 isotopes of, 46, 47 Lewis symbol for, 290 in living organisms, 58 organic compounds of. See Organic chemistry other group 4A elements vs., 950 oxides of, 946–48 thermodynamic quantities for, 1059–60 Carbon–11, 893 Carbon–14, 887–88, 892 Carbonated beverages, 524 Carbonate ion, 62, 63, 676 marine shell formation and, 728 Carbonates, 948 Carbon black, 946 Carbon-carbon bonds, 948, 950, 1007 in alkenes, 1015 in polymers, 491 rotation about, 1009 Carbon-carbon double bonds in lipids, 1037 rotation about, 1016 vulcanization of rubber and, 496 Carbon-carbon pi bond, 918 Carbon-carbon triple bonds, 1017 Carbon composites, 947 Carbon dioxide, 53, 384, 419, 442, 947 absorption of, by ocean, 766 acid-base reaction and, 129 atmospheric, 696, 701, 703–4, 751, 759, 761–64 global climate change and, 191 in blood, 713 bonding in, 919 climate change and, 192 critical temperature and pressure of, 442 as greenhouse gas, 421, 761–64 molecular model of, 4 nonpolarity of, 343 in oceans, 703–4 phase diagram of, 446–48 reaction with water, 267–68 in seawater, 765 specific heat of, 176 standard enthalpy of formation for, 184
supercritical, 442, 773 Carbon dioxide budget, 778 Carbon dioxide fire extinguisher, 396 Carbon disulfide, 554, 825, 949 Carbon fibers, 947 Carbon geometries, 1006 Carbon-hydrogen bond, 303, 1006, 1007 Carbonic acid, 129–30, 690, 713, 758, 766, 948 acid-dissociation constant of, 674 Carbonic acid-bicarbonate buffer system, 713 Carbonic acid buffer system, 728 Carbonic anhydrase, 605, 607–8, 1002 Carbon monoxide, 53, 380, 384, 758, 946 in atmosphere, 751, 758 in automobile exhaust, 592 from incomplete combustion, 83n reaction of iron oxide with, 648 standard enthalpy of formation for, 184 toxicity of, 1003 Carbon nanotubes, 498–99, 945 Carbon suboxide, 961 Carbon tetrachloride, 396, 522–23, 926 molal boiling-point elevation of and freezing-point depression constants of, 534 Carbonylation, 824, 1026 Carbonyl group, compounds with, 1024 aldehydes and ketones, 1024–25 amines and amides, 1028 carboxylic acids and esters, 1025–27 Carborundum (silicon carbide), 949 Carboxylate anion, 688 Carboxyl group, 688, 689 Carboxylic acids, 688–89, 1025–27, 1037, 1038 reaction with amine, 493–94 Carboxylic groups, 1007 Car exhaust gases, formaldehyde in, 778 Carnot, Sadi, 788, 824 Cars air bags, 83 battery, 855 emission standards for, 758 Carvone, 1025 Cassiopeia A, 874 Catalase, 592, 933 Catalysis, 559, 589–95 defined, 589 enzymes, 591–95 equilibria and, 637 in green chemistry, 772 heterogeneous, 590–91, 592 homogeneous, 589–90 Catalytic converters, 592, 761 Cathode, 258, 836–37 Cathode rays, electrons and, 41–43 Cathode-ray tubes, 41, 42 Cathodic protection, 859 Cation(s), 54 acidity of hydrated, 690–91 with anions, 683–85 chemical formulas and, 119 common, 60 in freshwater, 767 in ionic liquids, 436 from metal atoms, 59 names and formulas of, 59–60 from nonmetal atoms, 60
Index qualitative analysis to group, 737–38 reaction with water, 682–83 size of, 256, 257 Caustic soda (NaOH) acetic acid titration with, 716–20 hydrochloric acid titration with, 714–16 Cavendish, Henry, 920 Caves, limestone, 948 Cavities, tooth, 730 Cell emf, 838–45 concentration effects on, 849–54 equilibrium and, 850 free-energy change and, 847–49 oxidizing and reducing agents, 843–45 standard reduction (half-cell) potentials, 839–43 Cell membrane, lipids in, 1038, 1039 Cell metabolism, free energy and, 814 Cell phones, 463 Cell potential (Ecell), 838–45 under nonstandard conditions, 849–54 Cellulose, 20, 490, 603, 1036, 1037 Cellulosic plants, bioethanol from, 192 Celsius scale, 17 Cementite, 475 Cenotes, 114, 115, 116 Centered lattices, 467 Centimeter, 208 Centi prefix, 16 Cerium (Ce), 233 CERN (Conseil Européen pour la Recherche Nucléaire), 884 Cesium (Cs), 232, 265, 269 Cesium chloride, 482–83 Cesium ion (Cs+), 60 CFCs (chlorofluorocarbons), 457, 574, 756–57, 764 CH2 group, 21 Chadwick, James, 44 Chain reactions, 897 Chalcocite, 814 Chalcogens (group 6A), 51, 934–36 group trends for, 273–74 Challenger disaster, 173 Changes, 12–13. See also Reaction(s) of state, 12 Charcoal, 946 fuel value and composition of, 190 Charge(s) atomic, 41 conservation of, 123 electrical quantity of electrolysis and, 861–62 SI unit for, 42n electronic, 45 formal, 307–9 ionic, 55–56 in metal complexes, 972–74 partial, 309 Charged particles, accelerating, 884–85 Charge-transfer colors, 993 Charge-transfer transition, 993 Charles, Jacques, 390, 419 Charles’s law, 389–90, 391, 394 Chelate effect, 975 Chelating agents (polydentate ligands), 974–75, 998 in living systems, 976–79 Chemical analysis, 144–48
Chemical biology. See Biochemistry Chemical bond, 289. See also Bond(s) and bonding Chemical changes, 12–13. See also Reaction(s) Chemical energy, 162 Chemical equilibrium. See Equilibrium/equilibria Chemical industry, 6 Chemical kinetics, 558. See also Reaction rates Chemical properties, 11, 39 Chemical reactions. See Reaction(s) Chemicals, top eight produced, 6 Chemical structure, acid-base behavior and, 685–89 Chemiluminescence, 575 Chemistry atomic and molecular perspective of, 4–5 as central science, 5 chemical industry and, 6 defined, 3 descriptive, 917 study of, 4–6 reasons for, 5–6 Chemists, 6 Chile saltpeter, 937 Chiral, 983–84 Chirality in amino acids, 1030 in living systems, 1030 in organic chemistry, 1028–29 Chloral hydrate, 329 Chloramine, 938 Chlorate ion, 63 Chlorate salts, 929 Chloric acid, 125, 687 Chloride group, 738 Chloride ions, 63, 653 Chlorides insoluble, 737 in seawater, 765 Chlorine (Cl), 8, 274–75, 926, 927–28 added to municipal water supplies, 417–18, 769 bonding between tin and, 304 bonding in, 477 covalent bonds in, 345 electron affinity of, 263 Lewis symbol for, 290 mass spectrum of, 49 nonbonding electron pairs in, 477 nuclides of, 913 polarity of, 345 properties of, 275 reactions of with magnesium, 272 with methane, 316–17 with nitric oxide, 647 with ozone, 756–57 with phosphorus trichloride gas, 942 with sodium, 291–92 sale of, 6 state at room temperature and standard pressure, 427 uses of, 275 in water purification, 770–71 Chlorine–36, 915 Chlorine atoms, mass spectrum, 49
Chlorine dioxide, 423 water disinfection with, 771 Chlorine monoxide (ClO), 327, 757 3-Chlorobenzoic acid, 780 Chlorofluorocarbons (CFCs), 457, 574, 756–57, 764 Chloroform, 534, 555 Chloromethane, 1022 3-Chloroperbenzoic acid, 780 Chlorophyll a, 976 Chlorophylls, 976, 978 Chlorosis, 978 Chlorous acid, 667, 668, 687 Cholesteric liquid crystalline phase, 450 Cholesterol, 1023–24 Cholesteryl benzoate, 448 Chromate ion, 63, 993 Chromatography, 14 Chrome yellow, 993 Chromic or chromium(III) ion(Cr3+), 60 Chromium (Cr), 237, 265, 474, 968 electron configuration of, 237 oxidation in aqueous solution, 136 Chromium(III), 60, 973 Chromium(III) complexes, 989 Chromium oxides, 932 Chrysotile, 951 Cinnabar, 935 Cinnamon, 1025 cis fats, 1037 cis isomers, 969–70, 982 Cisplatin, 379, 982 Cis-trans isomerization, 380 Citric acid, 125, 127n, 651, 674, 697, 1025 acid-dissociation constant of, 674 Clausius, Rudolf, 402, 405 Clausius-Clapeyron equation, 444 Claus process, 111 Clausthalite, 507 Clean Air Act, 758 Click reaction, 775 Climate, 761–64, 766 Climate change, 191, 763 carbon dioxide and, 192 temperature of ocean water and, 781 Climate Change Conference (2009), 191 Closed system, 162, 771 Close-packed arrangement, 74 Close packing, 470–473 cubic, 470, 471, 472 hexagonal, 470, 471, 472 C60 molecules, 498–99 Coagulation, 544 Coal, 190–91 combustion of, 758 fuel value and composition of, 190 Cobalt, oxidation in aqueous solution, 136 Cobalt(II) chloride, 635 Cobalt(II) or cobaltous ion (Co2+), 60 Cobalt(III), 973, 990–91 ammonia complexes of, 969 Cobalt–60, 875, 885, 907, 912 Cobalt glass, 952 Cocaine, 701 Codeine, 680, 698 Coefficient, 84, 96 subscript vs., 79 Coffee-cup calorimeter, 177–78 CO group, 21 Cohesive forces, 438 Coke, 638
I-5
Colligative properties of solutions, 530–41 boiling-point elevation, 533–34 of electrolyte solutions, 540–41 freezing-point depression, 534–36 molar mass determination through, 539–40 osmosis, 536–39 vapor pressure reduction, 530–33 Collision model, 576 Collisions, molecular, 576 Colloidal dispersions, 541–42 Colloids, 541–46 hydrophilic and hydrophobic, 542–44 removal of colloidal particles, 544–46 types of, 541 Colors, 963 complementary, 244, 986 in coordination compounds, 985–87 charge-transfer, 993 crystal-field theory and, 987–88, 993 ligands and, 985 as function of temperature, 210 Color wheel, 986 Columbia Space Shuttle, 929 Combination reactions, 81–83 Combined gas law, 395 Combining volumes, law of, 390 Combustion, 1013–14 enthalpies of, 183 heat of, 183 Combustion analysis, 95–96 Combustion reactions, 76, 83–84 of alkanes, 1013–14 balanced equations for, 84 incomplete, 83n with oxygen, 919 oxygen as excess reactant in, 100 Common-ion effect, 704–7, 726–27 solubility and, 726–27 Common logarithms, 1053–54 significant figures and, 1054 Common names, 59, 1010 Complementary base pairs, 1040, 1041 Complementary colors, 244, 986 Complete ionic equation, 123 Complexes. See Metal complexes Complex ions, 968 defined, 732 formula of, 972–73 solubility and, 731–33 Components of mixture, 10 of solution, 513 Composition, 4 Compound(s), 8–10, 11. See also Organic chemistry with carbonyl group, 1024 aldehydes and ketones, 1024–25, 1047 amines and amides, 1028 carboxylic acids and esters, 1025–27 coordination. See Coordination compounds defined, 7 gases as, 384 interhalogen, 378 intermetallic, 475 ionic. See Ionic compounds of metals with nonmetals, 266
I-6
Index
Compound(s) (cont.) molecular, binary, 65–66 naming. See Nomenclature organic, 59, 66–67. See also Organic chemistry Compound semiconductors, 488 Computers, 463 Computer screens, 451 Concentrated, 526 Concentration(s), 139–43, 526–30 acid-base titration to determine, 145–48 conversion of units of, 528–30 defined, 139 dilution, 141–43 effects on cell emf, 849–54 concentration cells, 852–54 Nernst equation, 849–51 of electrolyte, 140 equilibrium, 632–33 free energy change and, 811–12 interconverting molarity, moles, and volume, 140–41 Le Châtelier’s principles and, 631 in mass percentage, 526–27 molality, 527–29 molarity, 139–40, 527–30 in mole fraction, 527–29 in parts per billion (ppb), 526–27 in parts per million (ppm), 526–27 percent ionization and, 672 in reactant or product, changes in, 632–33 reaction rates and, 558, 559, 563–69 change with time, 569–75 rate laws, 565–69 of solid substance, 623 Concentration cells, 852–54 “Concept links” feature, 30 Condensation, heat of, 439, 440 Condensation polymerization, 493–94 Condensation polymers, 493 Condensation reaction, 943, 1024 Condensed electron configurations, 231–32 Condensed phases, 426 Condensed structural formulas, 1009 for alkyl groups, 1011, 1012–13 for cycloalkanes, 1013 Conducting polymer, 500–501 Conduction band, 487 Conductivity, 116–17 electrical, 469 of metals, 478, 481 thermal, 469, 478 Cones, 357 retinal receptor, 245 Conjugate acid, 655 Conjugate acid-base pairs, 654–56, 679–80, 681 Conjugate base, 655 Conservation of energy, 786 Conservation of mass, law of (law of conservation of matter), 40, 78 Constant(s) acid-dissociation. See Aciddissociation constant (Ka) base-dissociation, 676, 679–81 decay, 888, 889 dissociation, 1062–63
equilibrium. See Equilibrium constant Faraday, 847 formation, 732 gas, 392 Henry’s law, 524 ion-product, 659 molal boiling-point-elevation, 534 molal freezing-point-depression, 535 Planck’s, 211 rate, 565, 567, 568, 576 Rydberg, 213–14 screening, 252 solubility-product. See under Solubility equilibria van der Waals, 412 Constant composition, law of (law of definite proportions), 10, 40 Constant-pressure calorimetry, 177–78 Constant-volume (bomb) calorimetry, 178–79 Constructive combination, 358, 364 Containment shell, 899 Continuous spectrum, 213 Contour representations, 223–24, 225–26 Control rods, 898 Convection, 180 Conversion factors, 25–28 involving volume, 28–29 using two or more, 27–28 Cooling, 440, 441 equilibria and, 635, 636 super–, 441 Cooling liquid, in nuclear reactor, 898–99 Coordination chemistry, 964 Coordination compounds, 964 colors in, 985–87 charge-transfer, 993 crystal-field theory, 987–95 colors and, 987–88, 993 d orbitals and, 988–89 electron configurations in octahedral complexes, 990–91 for tetrahedral and square-planar complexes, 991–95 defined, 968 isomerism in, 981–85 stereoisomerism, 981, 982–85 structural, 981 ligands, 968, 974–79 bidentate, 974 in living systems, 976–79 monodentate, 974 polydentate (chelating agents), 974–75, 998 weak field and strong field, 989 magnetism in, 987 metal complexes, 968–74 charges, coordination numbers, and geometries, 972–74 metal-ligand bond, 971–72 Werner’s theory and, 969–71 molecular-orbital theory of, 994 nomenclature of, 979–80 Coordination number, 470, 969 charges, geometries, and, 972–74 in ionic compounds, 482 Coordination sphere, 969, 970 Coordination-sphere isomers, 981 Copernicium (Cn), 886 Copolymers, 494 Copper (Cu), 8, 51, 469
in alloys, 473, 476 as electrical conductor, 113 electron configuration of, 237 extraction from chalcocite, 814 oxidation of, 131, 135–37 reaction with nitric acid, 12 Copper(I) or cuprous ion (Cu+), 60 Copper(II) or cupric ion (Cu2+), 60 Copper(II) sulfate, 10, 858n Copper(II) sulfate pentahydrate, 518 Coral reefs, 702, 703–4 Core electrons, 231 Corn, ethanol derived from, 20 Corrosion, 131, 857–59 of iron, 131, 858–59 as spontaneous process, 787 Coster, D., 246 Coulomb (C), 42n, 861 Coulomb’s law, 46 Covalent bond(s), 289, 290, 296–98, 304 bond enthalpies and strength of, 316–18 in hydrogen, 921 intermolecular forces vs., 427 octet rule exceptions, 312–14 orbital overlap and, 345–46 polar vs. nonpolar, 299 strengths of, 315–21 Covalent carbides, 949 Covalent-network solids, 464, 486–90 Covalent radius (bonding atomic radius), 254 Cracking, 923, 1014 Cracking catalysts, 590 CRC Handbook of Chemistry and Physics, 30 Crenation, 537–38 Creutz–Taube complex, 1001 Critical mass, 897 Critical point (C), 445, 446 Critical pressure, 441–42 Critical temperature, 441–42 Cross-linking of polymers, 495–96 Crustaceans, 728 Crutzen, Paul, 756 Cryolite, 862, 926 Crystal-field splitting energy, 988–89, 990 Crystal-field theory, 987–95 colors and, 987–88, 993 d orbitals and, 988–89 electron configurations in octahedral complexes, 990–91 for tetrahedral and square-planar complexes, 991–95 Crystal lattice, 465–67 Crystalline solids, 427, 465 entropy of, 799, 800 Crystallinity of polymers, 495 Crystallization, 518–19 Crystals ionic, 481, 483 liquid, 448–52 phases of, 448–49 properties of, 450 X-ray diffraction by, 468 Cubic centimeter, 18 Cubic close packing, 470, 471, 472 Cubic lattice, 466 Cupric or copper(II) ion (Cu2+), 60 Cuprous ion or copper(I) ion (Cu+), 60 Curie (Ci), 889
Curie, Marie and Pierre, 43 Curie temperature, 968, 1000 Curium–242, 886 Curl, Robert, 498 Current, electric, SI unit for, 15 Cyalume light sticks, 575 Cyanide ion, 63, 307 Cyanogen, 413 Cycles, 208 Cyclic glucose, 1034 Cycloalkanes, 1013 Cyclohexane, 75, 521, 550, 1013 Cyclohexanol, 75 Cyclohexatriene, 1020 Cyclopentadiene, 607 Cyclopentane, 1013 Cyclopropane, 422, 554, 1013 Cyclotrimethylenetrinitramine (RDX), 328 Cyclotron, 884 Cysteine, 1031 Cytochrome, 873 Cytoplasm, 1038 Cytosine, 459, 1040, 1041 Dalton, John, 40, 399 Dalton’s law of partial pressures, 399–400 Darmstadtium–273, 886 Data, 15 Dating, radiocarbon, 887–88 d-d transition, 989, 993 de Broglie, Louis, 216 Debyes (D), 302 Decane, 1009 viscosity of, 437 n-Decane, 509 deca- prefix, 65 Decarbonylation reaction, 822 Decay, radioactive, 877, 878–80 rates of, 886–91 types of, 878–80 Decay constant, 888, 889 Decimal places, in addition and subtraction, 24 Deci prefix, 16 Decomposition reactions, 82–83 Decompression sickness (the “bends”), 525 Deep-sea diving, blood gases and, 525 Definite proportions, law of (law of constant composition), 10 Degenerations, 226 Degradation, designing for after-use, 772 Degrees of freedom, 798, 799, 800, 801–2 Dehydration, 143, 936 Delocalization, 355–56, 1020 Delta (¢), 560 Democritus, 40 Density electron, 220 of gas, molar mass related to, 395–97 of liquid and solid phases, 433 probability, 220, 224 radial probability, 222 of seawater, 766 SI unit of, 19 weight vs., 19 Dental amalgam, 473 Deoxyhemoglobin, 977 Deoxyribonucleic acid (DNA), 1038 antiparallel strands, 1049 hydrogen bonds in, 433 replication of, 1040, 1041
Index Deoxyribose, 1040 Department of Energy, 52, 901 Dependent variable, 1056 Deposition, heat of, 439, 440 Derivatives, reduction of, 772 Derived SI units, 18 Desalination, 768–69 Descriptive chemistry, 917 Destructive combination, 358, 364 Detergents, 943–44, 1007 Deuteration, 921 Deuterium, 903, 920–21 Deuterium oxide, 696, 920 Dextrorotary, 984–85 Dextrose. See Glucose Diabetes, 90 Diacid, 494 Dialysis, 544 Diamagnetic solid, 967 Diamagnetism, 366–68 Diamine, 494 Diamond, 38 melting and boiling points of, 428 standard enthalpy of formation for, 184 structure of, 486–87 synthetic, 946 Diastolic pressure, 388 Diatomic gases, 384 Diatomic molecules, 53 bond enthalpies in, 315 dipole moments of, 302 electron configurations for, 365–66 heteronuclear, 369–71 homonuclear, 361–71 energy-level diagram for, 361, 365 molecular orbitals for, 361–71 Diazepam (valium), 330, 331–32 Diazine, 353 Diborane, 72, 203, 953 trans-2,3-Dichloro-2-butene, 1018 Dichlorobenzene, 376 Ortho-Dichlorobenzene, 327 Dichloroethylene, 376 1,2-Dichloroethylene, 459 Dichloromethane, 459 Dichromate ion, 63 Dicyclopentadiene, 607 Diethylether, 49, 201, 1024 vapor pressure of, 444 Differential rate law, 569 Diffraction, 217, 468 neutron, 243 Diffraction grating, 468 Diffusion, 405–9 mean free path and, 408–9 Difluoromethane, 373 Di-Gel, 130 Dilute, 526 Dilution, 141–43 Dimensional analysis, 25–30, 140, 393 conversion factors in, 25–28 conversions involving volume, 28–29 Dimers, hydrogen-bonded, 649 Dimethyl ether, 431, 1022 Dimethylhydrazine, 961 Dimethylphosphinoethane (dmpe), 1000 2,2-Dimethylpropane, 430, 1010 Dimethyl sulfide, 935 Dinitrobenzene, isomers of, 1021 Dinitrogen pentoxide, 608, 939
Dinitrogen tetroxide, 616, 867 Dinitrogen tetroxide-nitrogen dioxide equilibrium, 616, 621–22 Dinitrogen trioxide, 939 Dioxane, 550 Dioxygen, 273, 367–68, 930 Dipole, 301 Dipole-dipole forces, 428, 430–31, 433, 435 Dipole moments, 301–3, 343 di- prefix, 65 Diprotic acid, 124, 664 titration curve for, 720 Direct methanol fuel cells, 857 Disaccharides, 1035–36 Disilicate ion, 950 Disorder. See Entropy(ies) Dispersion forces, 429–30, 433, 435 solution formation and, 514–15 Displacement reactions, 133–34 Disproportionate reaction, 872, 940 Dissociation constants for acids, 1062 for bases, 1063 Dissociation energy, 753–54 Dissolution of ionic compounds, 722 in water, 116, 117 Distillation, 13–14, 532, 768 fractional, 1014 Division in exponential notation, 1052 significant figures in, 24 Djerassi, Carl, 930 DNA (deoxyribonucleic acid), 1038 antiparallel strands, 1049 hydrogen bonds in, 433 replication of, 1040, 1041 DNA double helix, 459 Dobson unit, 756 Dolomite, 948 Donor atom, 974 Dopants, 268, 489 Doping, 489–90 d orbitals, 225–26, 349–50, 354 crystal-field theory and, 988–89 phases in, 364 in square-planar complexes, 991–95 in tetrahedral complexes, 991–95 in transition metals, 965 Doses, radiation, 904–5 Double bonds, 298, 919, 1006 bond length of, 318–20 in retinal, 357 rotation about, 1016 Double helix, 1040–41 d-p backbonding, 381 Drugs amine, 680 anticancer, 379 chirality and, 1029 cisplatin, 379 lithium, 271 transplatin, 379 Dry ice, 947 Ductility, 469, 478 Dumas-bulb technique, 418–19 Dyes, 127, 146, 652 azo, 380 Dynamic equilibrium, 443, 519, 611 Dynamite, 319
E10, 20 E85, 20, 202 Earth, 418, 749 elements in crust, 8 surface temperature of, 761–62 thermal balance of, 761–62 water of, 764–71 freshwater and groundwater, 766–67 global water cycle, 764 human activities and, 767–71 salt water, 765–66 ECF (extracellular fluid), 853 Echinoderms, 728 Economy, atom, 771, 774–75 Edema, 538 EDTA (ethylenediaminetetraacetate ion), 974, 975 Effective nuclear charge (Zeff), 251–54 estimating, 253 for period 2 and period 3 elements, 253 for valence electrons, 252, 254 Effusion, 405–9 Graham’s law of, 407–8 Einstein, Albert, 211–12, 213, 242, 898 Eka-aluminum, 251 Eka-manganese, 280 eka- prefix, 280 Eka-silicon, 251, 280 Elastomers, 492, 825 Electrical charge quantity of electrolysis and, 861–62 SI unit for, 42n Electrical circuit, 836 Electrical conductivity, 469 Electrical work, 172, 849 Electric current, SI unit for, 15 Electricity from nuclear fission, 899, 901 from nuclear reactions, 876 Electrocardiography, 853 Electrochemistry, 827–73. See also Oxidation-reduction (redox) reactions batteries, 826, 827, 854–57 alkaline, 855–56 fuel cells, 856–57 lead-acid, 855 nickel-cadmium, nickel-metal hydride, and lithium-ion, 856 primary and secondary, 855 cell emf, 838–45 concentration effects on, 849–54 equilibrium and, 850 oxidizing and reducing agents, 843–45 standard reduction (half-cell) potentials, 839–43 corrosion, 857–59 of iron, 858–59 defined, 827 electrolysis, 860–63 with active electrodes, 860 of aqueous solutions, 860 electrical work, 849 quantitative aspects of, 861–62 of water, 9 free energy and redox reactions, 845–49 oxidation states, 828–29
I-7
voltaic (galvanic cells), 835–37 molecular view of electrode process, 865 work done by, 849 Electrode(s), 836 electrolysis with active, 860 molecular view of, 865 standard hydrogen, 839–40 Electrolysis. See under Electrochemistry Electrolyte(s), 117 colligative properties of, 540–41 common-ion effect and, 704 concentration of, 140 strong and weak, 118–19, 664 identifying, 126–27 in voltaic cell, 837 Electrolytic cells, 860 Electrolytic properties, of aqueous solution, 116–17 Electromagnetic force, 46 Electromagnetic radiation, 208 Electromagnetic spectrum, 208 Electrometallurgy of aluminum, 862 Electromotive force (emf). See Cell emf Electron(s), 44–45 bonding, 359 cathode rays and, 41–43 core, 231 discovery of, 42 mass of, 42–43 nonbonding, bond angles, and, 338–39 odd number of, 312 outer-shell, 231–32 paired, 229 unpaired, 229 valence, 231, 233–35, 236, 237 bonding and, 290 effective nuclear charge experienced by, 252, 254 as waves, 217 Electron affinity(ies), 263–64 electronegativity and, 299 of halogens, 264, 275 ionization energy vs., 263 of nonmetals, 267 Electron capture, 878, 879 Electron configuration(s), 229–33, 365–69 of actinides, 232–33 anomalous, 237–38 condensed, 231–32 homonuclear diatomic molecules, 361 Hund’s rule, 229–31 of ions, 262–63, 294–96 of lanthanides, 232–33 molecular properties and, 366–69 in octahedral complexes, 990–91 orbital diagrams and, 231 periodic table and, 233–38 of transition metals, 232, 965–67 Electron density, 220 in molecules, 296 in p orbitals, 224, 225 in s orbital, 222, 225 Electron-domain geometry, 335–38 Electron domains, 334–38 axial, 339 equatorial, 339 for multiple bonds, 334 Electronegativity bond lengths and, 303 bond polarity and, 300–301
I-8
Index
Electronegativity (cont.) ionic vs. covalent bonding and, 304 of oxyacids, 686–87 Electron–electron repulsion, 254 Electronic charge, 45 Electronics, ion movement in, 258 Electronic structure, 206–47 atomic orbitals, 219–22 d, 225–26 f, 225–26 p, 224–25 quantum numbers and, 220–22 s, 222–24 Bohr model and, 213–16 energy states of hydrogen atom, 214–16 limitations of, 216 defined, 207 line spectra and, 213–14 of many-electron atoms, 226–29 photons and, 210–12 quantization of energy and, 210–12 quantum (wave) mechanics, 207 wave behavior of matter and, 216–19 wave nature of light and, 208–10 Electron microscope, 217 Electron-pair acceptor/donor concept, 689–90 Electron-sea model for metallic bonding, 478 Electron shell, 220 Electron spin, 227–29 experimental evidence for, 227 Electron transfer, 280, 291 Electron volt, molar equivalent of, 488n Electroplating, 860 Electrostatic potential energy, 160–61 Element(s), 7–8, 11. See also specific elements and groups atoms and, 8 common, 8 defined, 4 discovery of, 250–51 electronegativities of, 299 molecules of, 4 nuclear synthesis of, 903 periodic properties of. See Periodic properties of elements relative abundances of, 8 representative (main-group), 233 required by organisms, 58 symbols for, 8 trace, 58 transuranium, 52, 885–86 Element 117, 285 Elemental form, atom in, 132 Elemental semiconductors, 488 Elementary reactions, 581 rate laws for, 583–84 emf. See Cell emf Emission spectra, 210 Emission standards, 758 Empirical formulas, 53, 92–96 calculating, 93–94 combustion analysis, 95–96 for ionic compound, 58–59 molecular formula from, 94 Emulsifying agent, 544 Enantiomers (optical isomers), 983–84, 1028–29, 1030
Endothermic processes, 167, 168, 171, 516–17, 635 End point of titration, 146n, 721 Energetics, of ionic bond formation, 292–94 Energy(ies), 159–64. See also Thermodynamics: first law of activation, 577–78 catalysis and, 593, 637, 638 determining, 579–81 biomass, 191 chemical, 162 conservation of, 786 crystal-field splitting, 988–89, 990 defined, 160 describing and calculating changes in, 163–64 dissociation, 753–54 electrostatic, 160–61 fossil fuels and, 190–91 geothermal, 191 global population growth and demand for, 192 hydroelectric, 191 internal, 164–65 ionization, 259–63, 299 of alkaline earth metals, 272 electron affinity vs., 263 of metals vs. nonmetals, 265 periodic trends in, 260–62 kinetic, 160–62 lattice, 292–95 magnitudes of, 293 molecular motion and, 796–97 nonrenewable, 191 nuclear, 191 nuclear binding, 895–96 of orbitals, 226, 370 of photon, 212 potential, 160–62 quantization of, 210–12 radiant, 208 renewable, 20–21, 191 solar, 20–21, 191–92, 370 sources of, in U.S., 190 spin-pairing, 990–91 Sun as source of, 875 system and surroundings, 162–63 thermal, 162 transferring, 163–64 units of, 162 wind, 191 Energy barrier, 577 Energy changes accompanying phase changes, 439–40 in nuclear reactions, 894–96 solution formation and, 515–17 Energy conversion, 814 Energy diagram, 165 Energy efficiency, designing for, 772 Energy Independence and Security Act of 2007, 20 Energy-level diagram (molecular orbital diagram), 359, 360 Energy Policy Act of 2005, 20 Energy states, of hydrogen, 214–16 Engine, ideal, 788 Enstatite, 950 Enthalpy(ies) (H), 169–75 bond, 315–21 bond length and, 318–20
enthalpies of reactions and, 316–18 change in (¢H), 170–72 sign of, 171–72 of combustion, 183 defined, 169 of formation, 183–87 to calculate enthalpies of reaction, 185–87 equations associated with, 184–85 free energy and, 808 of fusion, 183, 439 of reaction, 172–75, 185–87, 316–18 spontaneous processes and, 175 as state function, 169 of vaporization, 183 Enthalpy change Hess’s law and, 181–83 solution formation and, 516–17 Enthalpy diagram, 173 Hess’s law illustrated with, 183 Entropy(ies), 786, 790–803 absolute, 800, 801, 802 of activation, 825 biochemistry and, 1029 chelate effect and, 977 of expansion, 792 free energy and, 808 heat transfer and temperature related to, 790 life and, 800 microstates and, 794–96 molecular interpretation of, 793–800 probability and, 795 in reactions, 800–803 second law of thermodynamics, 790–93 solution formation and, 514, 515, 521 standard molar, 801–2 temperature and, 800–801 of universe, 793 Entropy change, 790–92 expansion and, 792 for phase changes, 791 qualitative predictions about, 797–99 in reactions, 800–803 in surroundings, 792–93, 802–3 of system, 792–93 Environment, chemistry of, 748–83 atmosphere, 750–64 composition of, 750–52 methyl bromide in, 574, 757 ozone in stratosphere, 754 temperature of, 750 troposphere, 750 Earth’s water, 764–71 freshwater and groundwater, 766–67, 768, 769–71 global water cycle, 764 human activities and, 767–71 salt water, 765–66 green chemistry, 771–75 atom economy, 774–75 principles of, 771–72 solvents and reagents, 773–75 Environment, radiation in, 902–8 Environmental Protection Agency (EPA), 906, 930, 945 Enzyme inhibitors, 593 Enzymes, 285, 591–95, 1036 efficiency of, 593
inhibition of, 593 in nitrogen fixation, 594 specificity of, 592–93 Enzyme-substrate complex, 593 EPA(Environmental Protection Agency), 906, 930, 945 Epinephrine (adrenaline), 108, 552, 698 Epsom salts, 109 Equation(s), 78–81 balanced, 78–80 for combination and decomposition reactions, 83 quantitative information from, 96–99 balancing, 78–80 direction of, 620–21 equilibrium constants and, 621–22 Henderson-Hasselbach, 709, 711 ionic, 122–24 molecular, 122–23 nuclear, 877–78 states of reactants and products, 81 thermochemical, 173 Equatorial electron domains, 339 Equilibrium constant, 613, 614–22 calculating, 625–26, 632–33 direction of chemical equation and, 620–21 evaluating, 616–17 Gibbs free energy and, 811–15 magnitude of, 619–20 in terms of pressure, 617–18 thermodynamic, 618 units of, 618 Equilibrium-constant expression (equilibrium expression), 614–15 Equilibrium constants, 1062–63 Equilibrium/equilibria, 119, 610–49 acid-base. See Acid-base equilibria aqueous. See Aqueous equilibria cell emf and, 850 chemical, 119 concept of, 612–14 dynamic, 443 free energy and, 804–5 heterogeneous, 623–25 homogeneous, 623 Le Châtelier’s principle, 631–37, 704 catalyst effects, 637–40 change in reactant or product concentrations, 632–33 nitric oxide emission control and, 640 temperature changes, 634–37 volume and pressure changes, 633–34 static, 611 Equilibrium mixture, 612–13 Equivalence point of titration, 145, 714, 716, 717, 718, 719–20, 721 pH and, 716, 718–20 Eskalith, 271 Essential amino acids, 1030 Essential fatty acids, 1038 Esters, 509, 1025–27 Estimating answers, 26 Ethanal (acetaldehyde), 1022, 1024 Ethanamide (acetamide), 1022, 1028 Ethane, 66, 67, 1008, 1009 bond enthalpies in, 315
Index combustion of, 1013 formation of, 591 molar entropies of, 802 in natural gas, 190 standard enthalpy of formation for, 184 1,2-Ethanediol (ethylene glycol). See Ethylene glycol Ethanoic acid. See Acetic acid (ethanoic acid) Ethanol (ethyl alcohol), 4–5, 20, 66, 67, 158, 457, 1014, 1023, 1025 as biofuel, 192 density of, 19 molar boiling-point elevation and freezing-point depression constants of, 534 molecular model of, 4 solubility of, 521 standard enthalpy of formation for, 184 vapor pressure of, 442, 444 Ethene (ethylene), 53, 384, 591, 1015, 1022 Ethers, 1024 Ethyl acetate, 377, 1025, 1026 Ethyl alcohol. See Ethanol (ethyl alcohol) Ethylamine, 1022, 1028 Ethylammonium nitrate, 436 Ethylbenzene, 1021 in styrene manufacture, 772 Ethyl butyrate, 108 Ethyl chloride, 457 Ethylene (ethene), 53, 1008, 1014, 1015, 1022 carbon-carbon double bond in, 357 empirical formula for, 53 hybridization in, 352 molecular formula for, 53 molecular geometry of, 352 orbital structure of, 353 pi bonds in, 352–53 reactions of with halogens, 1017–18 with hydrogen gas, 591 sale of, 6 standard enthalpy of formation for, 184 Ethylenediamine (en), 975, 977 Ethylenediaminetetraacetate ion (EDTA), 974, 975 Ethylene glycol, 5, 94, 455, 459, 530, 535, 1023 density of, 19 molecular model of, 4 vapor pressure of, 444 Ethylene glycol dimethyl ether, 455 Ethylene oxide, 554 Ethyl group, 591, 1011 Ethyl methyl ether, 455 Ethyne (acetylene), 353–54, 419, 822, 949, 1008, 1017, 1022 Eucalyptus, 112 Eutrophication, 768 Evaporation, 180 Exact numbers, 20 Excess reactants (reagents), 100 Exchange (metathesis) reactions, 121–22 Excited state, 214, 221 Exothermic processes, 167, 168, 171, 635 solution formation and, 516–17 spontaneity and, 516–17
Explosive balance, 319 Explosives, 319, 940 Exponential notation, 1051–53 Exponents negative, 1051 positive, 1051 in rate law, 565–67 Extensive properties, 11 Extracellular fluid (ECF), 853 ExxonMobil Research and Engineering Company, 20 Face-centered cubic unit cell, 467, 469, 470, 482 Faces (of solids), 465 Faceting, in ionic crystals, 481 Fahrenheit scale, 17 Faraday (F), 847 Faraday, Michael, 498, 553, 847 Faraday’s constant, 847 Fast breeder reactor, 900–901 Fats, 1037–38 metabolism of, 188 Fat-soluble vitamins, 522 Fatty acids, 1037–38 f-block metals, 234 FDA, 271 Feedstocks, renewable, 772 FeMo-cofactor, 594 Femto prefix, 16 Fe2O3, 59 Fe3O4, 59 Fermi, Enrico, 898 Fermium–253, 886 Ferrichrome, 978, 979 Ferric or iron(III) ion (Fe3+), 60, 690, 971 Ferrimagnetism, 968 Ferrochrome, 474 Ferromagnetism, 967 Ferrous or iron(II) ion (Fe2+), 60 Ferrous sulfate, 780 Fertilizers, 614, 615 nitrogen, 937 phosphorus, 943–44 Fibers, 947 Fibrous proteins, 1034 Film, 891 Film badges, 891 Filtration, 13 Fire extinguisher, carbon dioxide, 396 Fireflies, 5, 575 Fireworks, 272 First ionization energy, 259, 260–62 of alkaline earth metals, 272 of metals vs. nonmetals, 265 periodic trends in, 260–62 First law of thermodynamics, 164–69 algebraic expression of, 166 endothermic and exothermic processes, 167, 168 heat and work related to internal energy changes, 165–67 internal energy, 164–65 state functions, 167–69 First-order reactions, 565, 569–71, 574 half-life of, 573–75 Fission, nuclear, 896–901 reactors using, 898–900 waste from, 900–901 Fixed nitrogen, 937 Flame tests, 270
Flex-fuel vehicles, 202 Fluorapatite, 926 Fluorescence, 41 Fluorescent lamps, compact, 21 Fluoridation, 730, 747 Fluoride, in seawater, 765 Fluoride ion, 63 Fluorine (F), 8, 274–75, 276, 925, 926 electronegativity of, 300 Lewis symbol for, 290 nonpolarity of, 344 oxidation number of, 132 properties of, 275 thermodynamic quantities for, 1060 Fluorine–18, 893 Fluorite, 508 Fluoroapatite, 730, 746 Fluorocarbons, 554 Fluorspar, 926 Folding, 1033 Food and Drug Administration (FDA), 271 Foods, thermochemistry of, 188–90 “Food versus fuel” debate, 192 Fool’s gold (pyrite), 465, 935 f orbitals, 225–26 Force(s), 46, 160 adhesive, 438 cohesive, 438 defined, 163 dipole-dipole, 428, 430–31, 433, 435 electromagnetic, 46 electromotive. See Cell emf gravitational, 46 intermolecular. See Intermolecular forces ion-dipole, 428, 434, 435 nuclear, 46 Formal charge, 307–9 Formaldehyde (methanal), 354, 930, 1006, 1024 in car exhaust gases, 778 Formation constant (Kf), 732 Formic acid (methanoic acid), 150, 688, 746, 1025 acid-dissociation constant of, 668–69 decomposition of, 595 percent ionization of, 669 Formula(s), 52–53 condensed structural, 1009, 1011, 1012–13 empirical, 53, 92–96 calculating, 93–94 combustion analysis, 95–96 for ionic compound, 58–59 molecular formula from, 94 molecular, 53, 54 from empirical formulas, 94 percentage composition from, 93 relating relative numbers of anions and cations to, 119 structural, 54 subscripts in, 79, 84 Formula units, 85 interconverting mass and, 91 Formula weights, 84–86 molar mass and, 88–89 percentage composition from, 85–86 Fossil fuels, 190–91, 370 global population growth and demand for, 192
I-9
greenhouse effect and combustion of, 762 Fractional distillation, 1014 Fraunhofer lines, 245 Free energy, 803–8 equilibrium constant and, 811–15 under nonstandard conditions, 811–12 redox reactions and, 845–49 spontaneity and, 803–4 standard free-energy changes, 806 temperature and, 809–10 Free-energy change emf and, 847–49 Gibbs free energy and, 808 Free radical, 904 Freezing, heat of, 439, 440 Freezing point, 445 Freezing-point depression, 534–36 molar mass from, 539 Frequency, 208, 209–10 calculating from wavelength, 210 Frequency factor, 578 Freshwater, 766–67 dissolved oxygen and quality of, 768 pH of, 758, 759 softening of, 770 treatment of municipal supplies of, 769–71 Friedel-Crafts reaction, 1021 Frisch, Otto, 898 Fructose, 1034–35 Fuel cells, 856–57 direct methanol, 857 Fuel cell stack, 857 Fuel efficiency, 20 Fuel elements, 898 Fuel oil, 1014 Fuels, 162 fossil, 190–91, 370, 762 global population growth and demand for, 192 thermochemistry of, 190–91 Fuel source, sustainable, 158 Fuel values, 188 Fuller, R. Buckminster, 499 Fullerenes, 498–99, 945 Functional groups, 66–67, 1007, 1021–28 alcohols, 1023 aldehydes and ketones, 1024–25 amines and amides, 1028 carboxylic acids and esters, 1025–27 ethers, 1024 Furchgott, Robert F., 941 Furoic acid, 746 Fusion, 439 enthalpies of, 183 heat (enthalpy) of, 439 nuclear, 896, 902 Galactose, 1048 Galena, 935 Gallium, 249, 250–51, 265, 458 electron configuration of, 233 Gallium arsenide (GaAs), 488 Gallstones, 1024 Galvani, Luigi, 853 Galvanic cell, 835–37 Galvanic (voltaic) cells, 835–37. See also Batteries concentration cells, 852–54 electromotive force (emf) in, 838–45
I-10
Index
Galvanic (voltaic) cells (cont.) concentration effects on, 849–54 equilibrium and, 850 oxidizing and reducing agents, 843–45 standard reduction (half-cell) potentials, 839–43 molecular view of electrode process, 865 standard cell potential of, 839, 842 work done by, 849 Galvanized iron, 859 Gamma (g) radiation, 43, 208, 209, 878, 902, 904 therapeutic, 907 Gas(es), 7, 382–423 absolute temperature of, 403 acid-base reactions with, 129–30 blood, deep-sea diving and, 525 characteristics of, 384 collecting over water, 401–2 diatomic, 384 diffusion and mean free path, 408–9 entropy change and isothermal expansion of, 792 expansion of, 786, 787, 792 at molecular level, 793–94 in freshwater, 767 Graham’s law of effusion, 407–8 greenhouse, 761–64 ideal, 392, 533 ideal-gas equation, 391–95 gas density-molar mass relationship and, 395–97 gas laws and, 394–95 gas volume in reactions and, 397–99 inert, 276 isothermal expansion of, 792 kinetic-molecular theory of, 402–5 gas laws and, 404–5 mixtures of, 384, 399–402 monoatomic, 384 natural, 190, 255, 398, 423, 781, 824. See also Methane combustion of, 781 conversion of, 84 fuel value and composition of, 190 noble, 51, 924–25 boiling points of, 429 electron affinities for, 264 group trends for, 276–77 pressure, 385–87 partial, 399–401 pressure-volume work and, 170–71, 172 properties of, 384, 426 real, 409–13 van der Waals equation, 411–13 separations of, 408 solids and liquids compared to, 384 solubility of, 520, 523–26 state of, 387 water, 922 Gas constant, 392 Gas-cooled reactor, 899 Gas laws, 387–91 Avogadro’s law (quantity-volume relationship), 390–91 Boyle’s law (pressure-volume relationship), 388–89, 391
Charles’s law (temperature-volume relationship), 389–90, 391 combined, 395 ideal-gas equation and, 391–95 kinetic molecular theory and, 404–5 Gasoline, 521, 532, 1014 blends, 20 combustion of, 808 cracking catalysis to form, 590 fuel value and composition of, 190 Gas pipelines, 398 Gay-Lussac, Joseph Louis, 390 Geiger counter, 891–92 Geim, Andre, 500 Geometric isomerism, 982, 1015 Geothermal energy, 191 Gerlach, Walter, 227 Germanium (Ge), 250–51, 265, 486, 487, 488, 510, 950, 960 Gibbs, Josiah Willard, 803 Gibbs free energy (G), 803–8 equilibrium constant and, 811–15 under nonstandard conditions, 811–12 spontaneity and, 803–4 standard free-energy changes, 806 temperature and, 809–10 Gibbs free-energy change, 808 emf and, 847–49 Giga prefix, 16 Glacial acetic acid, 153 Glass, 465, 952 quartz, 465 volcanic (obsidian), 465 Glassware, volumetric, 18–19 Global climate changes See Climate change Global warming, 763. See also Climate change Globular proteins, 976, 1034 Glucose, 72, 89, 1007, 1034–35 cyclic, 1034 ethanol from, 192 as food, 188 molecular formula of, 1034 oxidation of, 98, 814 from photosynthesis, 193 solubility of, 521 standard enthalpy of formation for, 184 structure of, 1034 Glucose monitoring, 90 Glutamic acid, 651, 1031 Glutamine, 1031 Glutathione, 1050 Glycerin, 531–32 Glycerol (1,2,3-propanetriol), 153, 1023, 1037, 1038 Glycine, 377, 689, 700, 1030, 1031, 1032 anion of, 1002 Glycogen, 1036, 1037 Glycylalanine, 1030, 1031 “Go Figure” feature, 30 Goiter, 928 Gold, 51, 138, 779 Gold density of, 19 discovery of, 250 on nanoscale, 498 oxidation in aqueous solution, 136 oxidation states of, 872 properties of, 138, 498 reactions of, 138
Gold(III), 973 Gold alloys, 473, 474, 476–77 Gold leaf, 469 Goodyear, Charles, 496 Goodyear blimp, 417 Goudsmit, Samuel, 227 Graduated cylinder, 18, 19 Graham, Thomas, 407 Graham’s law of effusion, 407–8 Gram (g), 16 Grams, converting to moles, 90 Granite, 10 Graphene, 467, 498–500 Graphite, 945–46 as anode, 258 in batteries, 856 in carbon fibers, 947 structure of, 486, 487 Graphs, 1056 Gravitational forces, 46, 385 Gravity, 160–61 Gray (Gy), 904 Great Barrier Reef, 702, 703 Greek prefixes, 65, 980 Green chemistry, 771–75 atom economy, 774–75 principles of, 771–72 solvents and reagents, 773–75 Greenhouse effect, 762, 764 carbon dioxide and, 421, 762 methane and, 764 Greenhouse gases, 761–64. See also specific gases Grigg, C. L., 271 Ground state, 214, 220, 221 Group 1A elements. See Alkali (group 1A) metals Group 2A elements. See Alkaline earth (group 2A) metals Group 4A elements, 949–52. See also Carbon (C); Silicon (Si) general characteristics of, 949–50 Group 5A elements, 941–45. See also Nitrogen (N); Phosphorus (P) electron affinities of, 264 general characteristics of, 941–42 Group 6A elements (chalcogens), 51, 934–36. See also Oxygen (O) general characteristics of, 934 group trends for, 273–74 occurrences and production of, 934 properties and uses of, 934–35 Group 7A elements. See Halogens Group 8A (noble gases), 924–25 Groups, 50, 51 Guanine, 459, 1040, 1041 Guldberg, Cato Maximilian, 614 Guy-Lussac, Joseph Louis, 419 H+ ions concentrations of pH values and, 660–61 rate law and, 661 in water, 652–53 Haber, Fritz, 295, 615, 630–31, 637–38 Haber (Haber-Bosch) process, 614, 615, 873 free energy changes in, 810, 813 hydrogen and, 923 nitrogen and, 937 temperature effects on, 615, 631
Hafnium, 246, 965 Hahn, Otto, 898 Half-cell potentials, 839–43, 1064 Half-life of reactions, 573–75, 886–87 calculations based on, 887–91 Half-reaction, 830, 860–61 Half-reaction method, 830–33 Halides halogens and, 276 hydrogen, 928–29 phosphorus, 942 Hall, Charles M., 862 Hall-Héroult process, 862 Halogen-containing molecules, ozone layer destruction and, 574 Halogens, 51, 235, 926–30 boiling points of, 429 electron affinities of, 264 elemental, 275 group trends for, 274–76 hydrogen halides, 928–29 interhalogen compounds, 929 oxidation number of, 132 as oxidizing agents, 843 oxyacids and oxyanions of, 929 properties and production of, 926–27 reaction with ethylene, 1017–18 uses of, 927–28 Halons, 781 Handedness. See Chirality Hard drive, computer, 462–63 Hard water, 770 HDPE (high-density polyethylene), 494, 495 Heart ion concentration and, 853 nitroglycerin and, 941 Heat, 160. See also Enthalpy(ies) (H) of combustion, 183 condensation, 439, 440 of deposition, 439, 440 of formation, 183 of freezing, 439, 440 of fusion, 439 internal energy change and, 165–67 of reaction, 172–74 reversible flow of, 789 sign conventions for, 166 specific, 175–77 of sublimation, 439 transferring energy and, 163–64 of vaporization, 439 Heat capacity, 175–77 molar, 175 Heat exhaustion, 180 Heating curves, 440–41 Heat packs, 516 Heat stroke, 180 Heavy water, 920, 921 Heavy water reactor, 899 Heisenberg, Werner, 217–18 Helium (He), 8, 49, 51, 924 in atmosphere, 751 burning of, 903 deep-sea diving and, 525 electron configuration of, 230 formation of, 903 properties of, 276 Helium–4, 895 Helium ion (He2+), bond order of, 360 Heme, 976–77
Index Hemoglobin, 111, 545, 649, 699, 713, 976–77 Hemolysis, 537–38 Henderson-Hasselbach equation, 709, 711 Henry’s law, 524 Henry’s law constant, 524 Heptane, 1009, 1014 viscosity of, 437 hepta- prefix, 65 Héroult, Paul, 862 Hertz (Hz), 209 Hess’s law, 181–83, 316, 621 enthalpy diagram illustrating, 183 Heterogeneous alloys, 475 Heterogeneous catalysis, 590–91, 592 Heterogeneous equilibria, 623–25 Heterogeneous mixture, 10, 11 Heteronuclear diatomic molecules, 369–71 Hevea brasiliensis, 496 Hexafluorobenzene, 459 Hexafluorosilicic acid, 929 Hexagonal close packing, 470, 471, 472 Hexagonal lattice, 465, 466 Hexane, 1009 solubility of, 521 viscosity of, 437 Hexanol, 521 hexa- prefix, 65 Hexatriene, 312 HFCs (hydrofluorocarbons), 757, 764 High-carbon steels, 474 High-density polyethylene (HDPE), 494, 495 Highest occupied molecular orbital (HOMO), 370 High-spin complex, 991 High-temperature pebble-bed reactor, 900 Hindenburg disaster, 173 Histidine, 1031 Hoffmann, Roald, 930 Holes, 490 HOMO (highest occupied molecular orbital), 370 Homogeneous catalysis, 589–90 Homogeneous equilibria, 623 Homogeneous mixture, 10, 11. See also Solution(s) Homonuclear diatomic molecules, 361–71 Hong Kong skyline, 784 Hot-air balloons, 390 Household chemicals, 6 Human body elements in, 8 temperature regulation in, 180 Hund’s rule, 229–31, 232 Hurricane Wilma of 2005, 416 Hybridization, 346 Hybrid orbitals molecular geometry and, 346–51 involving d orbitals, 348–49 sp, 346–48 sp2 and sp3, 348–50 triple bonds and, 353–54 Hydrates, 518, 858n Hydration, 515, 690–91 water of, 518 Hydrazine, 867, 938–39, 960–61 Hydrazobenzene, 380 Hydrazoic acid, 697 Hydride ions, 63, 269, 921, 923–24 Hydrides, 269
binary, 686 interstitial, 924 metallic, 924 molecular, 924 Hydrobromic acid, 125 Hydrocarbon(s), 66–67, 1008–14 as atmospheric pollutants, 758 branched-chain, 1009 combustion of, 83–84 derivatives of, 66–67 immiscibility of, 521 as pollutants, 761 saturated (alkanes), 66–67, 1008, 1009–14 cycloalkanes, 1013 nomenclature of, 1010–13 reactions of, 1013–14 structural isomers of, 1009–10 structures of, 1009 straight-chain, 1009 structural and molecular formulas for, 67 unburned, 592 unsaturated alkenes, 1015–17 alkynes, 1017–19 aromatic, 1008, 1019–21 viscosities of, 437 Hydrocarbon fractions, from petroleum, 1014 Hydrochloric acid, 124, 125, 128, 129–30, 237, 275, 561 covalent bonds in, 345 ionization of solution of, 118 reactions of with cobalt(II) chloride, 635 with magnesium, 134 with nickel, 517–18 with sodium hydroxide, 127–28, 652 with zinc, 828 titration with caustic soda (NaOH), 714–16 Hydrocyanic acid, 667, 949 Hydroelectric energy, 191 Hydrofluoric acid, 109, 125, 929 properties of, 667 Hydrofluorocarbons (HFCs), 757, 764 Hydrogen (H), 8, 9, 51, 53, 273, 419, 920–24 abundance of, 2 activity series of, 136 in atmosphere, 751 atomic emission of, 213 binary compounds of, 923–24 combustion of, 173 covalent bonds in, 296, 345 electron configuration of, 229 energy levels in, 221 energy states of, 214–16 formation of, 903 fuel value and composition of, 190 group trends for, 273 ionization energy of, 273 isotopes of, 420, 920–21 line spectrum of, 213 metallic, 283 molecular, 53, 923 attractions and repulsions in, 296 covalent bonding in, 296 molecular orbitals in, 358–60
reaction with oxygen, 857 as reducing agent, 844 nuclear spin in, 228 orbitals of, 220, 221, 358–60 oxidation number of, 132 probability density in s orbitals of, 225 production of, 922–23 properties of, 9, 284, 921–22 reactions of, 273 with ethylene, 591 with nonmetals, 273 with oxygen, 173 Schrodinger’s equation for, 219–20 thermodynamic quantities for, 1060 uses of, 923 Hydrogenation, 1018, 1037 Hydrogen bomb, 902n Hydrogen bonding, 431–34 aqueous solubility and, 521 between complementary base pairs, 1040, 1041 in DNA, 1041 solution formation and, 515, 517 trends in, 431–32 in water, 432, 433–34 Hydrogen bromide, standard enthalpy of formation for, 184 Hydrogen burning, 903 Hydrogen carbonates, 948 Hydrogen chloride, 316–17, 652 melting and boiling points of, 428 standard enthalpy of formation for, 184 Hydrogen compounds of nitrogen, 937–39 Hydrogen cyanide, 113, 384, 949 Hydrogen economy, 327, 922 Hydrogen fluoride melting and boiling points of, 428 standard enthalpy of formation for, 184 Hydrogen fuel cells, 856–57 Hydrogen halides, 276, 928–29 bond lengths and dipole moments of, 303 charge separation in, 303 Hydrogen iodide, standard enthalpy of formation for, 184 Hydrogen ion, 60, 124, 125 Hydrogen oxalate ion, 695 Hydrogen-PEM fuel cell, 857 Hydrogen peroxide, 53, 274, 933, 1024 decomposition of, 592 reaction with bromide ions, 589–90 structural formula for, 54 Hydrogen sulfates, 936 Hydrogen sulfide, 111, 129, 384, 606, 935 critical temperature and pressure of, 442 Hydrogen sulfites, 656, 935 Hydroiodic acid, 125 Hydrolysis, 681 of esters, 1026–27 Hydronium ions, 132, 653, 682, 683 Hydrophilic colloids, 542–44 Hydrophobic colloids, 542–44 Hydroquinone, 773–74 Hydrosulfide ion, 676 Hydroxide ions, 63, 125 Hydroxides of alkali metals, ionic, 665–66 of alkaline earth metals, ionic, 665–66
I-11
amphoteric, 733–34 base-insoluble, 738 ionic, 665–66 metal, 125 solubility of, 733–34 Hydroxyapatite, 730, 746 Hydroxylamine, 676, 678, 938 Hydroxyl group, 1023 Hydroxyl radical, 781, 904, 913 Hypertension, 388 Hypertonic solutions, 537 Hypervalent molecules, 313–14 Hypobromous acid, 687 Hypochlorite ion, 676, 698 Hypochlorite salts, 929 Hypochlorous acid, 275, 667, 678, 687 Hypoiodous acid, 687 Hyponatremia, 143 hypo- prefix, 61, 64 Hypothalamus, 180 Hypothermia, 180 Hypothesis (tentative explanation), 15 Hypotonic solutions, 537 Ibuprofen, 108, 550 (S)-Ibuprofen, 1029 Ice, 7, 433–34 melting of, 175, 787, 792–93 structure of, 797 Ice packs, 516, 517 ICF (intracellular fluid), 853 -ic suffix, 60, 64 Ideal bond angles, 336 Ideal engine, 788 Ideal gas defined, 392 entropy and, 792, 796 Raoult’s law and, 533 Ideal-gas equation, 391–95, 405 gas densities and molar mass, 395–97 gas laws and, 394–95 gas volumes in chemical reactions, 397–99 Ideal solutions, 532 -ide suffix, 61, 62, 65 Ignarro, Louis J., 941 Imaging, medical, 228 Immiscible liquids, 521 Incandescent lightbulbs, 21 Incomplete combustion, 83n Independent variable, 1056 Indicators, acid-base, 145–46 titrating with, 721–22 Indigo, 1049 Indium alloyed with gold, 477 electron configuration of, 233 Indole, 1049–50 Inert gases, 276 Inexact numbers, 20–21 Information, tabulating, 393 Infrared radiation, 208, 209 Inorganic compounds, 59 nomenclature of, 59–66 acids, 64–65 binary molecular compounds, 65–66 ionic compounds, 59–66 Insoluble chlorides, 737 Insoluble phosphates, 738 Instantaneous reaction rate, 561–62
I-12
Index
Insulin, 90 Integrated rate law, 569–70, 571 Intensity, luminous, 15 Intensive properties, 11 Interhalogen compounds, 378, 929 Intermediates, 582–83 Intermetallic compounds, 475 Intermolecular forces, 425–61 attractive, 410, 411 covalent bond vs., 427 effect on gas pressure, 410, 411 flowchart for determining, 435 in gases, 427 ion-dipole, 428 in liquids, 427 molecular geometry and, 430 pressure and, 427–28 Raoult’s law and, 533 in smectic phases, 449 in solids, 427 in solution formation, 514–15 surface tension and, 437 van der Waals forces comparison of, 435 dipole-dipole, 428 hydrogen bonding, 431–34 London dispersion, 429–30 Internal energy (E), 164–65 change in, 164–65 relating to heat and work, 165–67 sign conventions for, 166 as state function, 167–69 International Union of Chemistry, 1010 International Union of Pure and Applied Chemistry (IUPAC), 51, 1011 Internuclear axis, 351 Interstitial alloys, 474, 475 Interstitial carbides, 949 Interstitial hydrides, 924 Intracellular fluid (ICF), 853 Invert sugar, 1036 Iodide ion, 63 Iodine (I), 8, 274–75, 926, 927, 928 bonding atomic radius of, 254 properties of, 275 reaction with methane, 646 state at room temperature and standard pressure, 427 thermodynamic quantities for, 1060 Iodine–131, 893, 907, 912 Iodine bromide, 378 Iodine pentafluoride, 423 Iodized salt, 928 1-Iodododecane, 416 Ion(s), 54–56 calcium, 58 carbonate, 676 charges of, 55–56, 292–93, 295–96 chemical symbols for, 55 complex, 968 electron configurations of, 262–63 formation of complex, 731–33 in human heart, 853 hydride, 269 hydrogen, 124, 125 hydroxide, 125 isoelectric series of, 258 monatomic, 132 negative. See Anion(s) oxide, 274 peroxide, 274
polyatomic, 55, 61, 306–7 positive. See Cation(s) precipitation and separation of, 734–36 properties of, 55 required by organisms, 58 sizes of, 254–59 spectator, 123, 124 sulfide, 736 superoxide, 270, 274 transition-metal, 296 Ion concentration, measuring using conductivity, 116 Ion-dipole forces, 428, 434, 435 solution formation and, 514–15 Ion-dipole interaction, 987 Ion-exchange column, 777 Ion exchange for water softening, 770 Ionic bonds, 289, 290, 291–96, 300, 304, 435, 481 electron configurations of ions and, 294–96 energetics of formation, 292–94 polyatomic ions, 300 transition-metal ions, 296 Ionic carbides, 948 Ionic compounds, 56–57 coordination numbers in, 482 dissolution or precipitation of, 722 electrolytic behavior of, 126 electrolytic strength of, 118, 119 formation of, 57, 291–92 lattice energies for, 292–95 names and formulas of, 59–64 solubilities of, 120–21, 525, 729 in water, 117–18 Ionic equation(s), 122–24 complete, 123 net, 123 writing, 123–24 Ionic hydrides, 923–24 Ionic liquids, 436 Ionic radii, 254–59 periodic trends in, 256–59 Ionic solids, 464, 481–85 empirical formula and density of, 484–85 properties of, 481 structures of, 482–85 in water, 516, 798 Ionization percent, 669, 671–73 photoionization, 752, 754 of weak base, 706 Ionization energy, 259–63, 299 of alkaline earth elements, 272 electron affinity vs., 263 electronegativity and, 299 of metals, 265 periodic trends in, 259–63 Ionizing radiation, 902 Ion pair, 540–41 Ion-product constant, 659, 679 Ion product of water, 659–60 Iridium, 506 Iridium–192, 907 Iron (Fe), 8, 510 corrosion of, 131, 858–59 density of, 19 galvanized, 859 in myoglobin and hemoglobin, 978
oxidation of, 131, 136, 787 as reducing agent, 844 specific heat of, 176 thermodynamic quantities for, 1060 Iron(II) fluoride, 863 Iron(II) or ferrous ion (Fe2+), 60 Iron(II) sulfide, 935 Iron(III), 973 Iron(III) chloride hexahydrate, 518 Iron(III) or ferric ion (Fe3+), 60 Iron(III) oxide, 858, 932 Iron–58, 885 Iron–59, 893 Iron oxide, reaction with carbon monoxide, 648 Iron pyrite (fool’s gold), 465, 935 Irreversible process, 788–90 Isoamyl acetate pheromone, 1005, 1006 Isobutane (2-methylpropane), 1010 Isoelectronic series of ions, 258 Isolated system, 163 Isoleucine, 1031 Isomerism, 67, 328, 818, 981–85 alkane, 1009–10 cis-trans, 380 coordination-sphere, 981 drawing, 1016–17 geometric, 982, 1015 linkage, 981 optical, 983–84 stereoisomerism, 981, 982–85 structural, 67, 981 Isooctane (2,2,4-trimethylpentane), 1014 Isopentane (2-methylbutane), 1010 Iso- prefix, 1010 Isoprene, 496 Isopropyl alcohol (2-propanol), 67, 95–96, 455, 1023 Isopropyl group, 1011 Isotactic polypropylene, 511 Isothermal process, 791, 792 Isotonic solutions, 537 Isotopes, 46–47, 876 abundance of, 48 of hydrogen, 920–21 stable, with even and odd numbers of protons and neutrons, 882–83 synthetic, 885, 886, 887 -ite suffix, 61, 62 -ium suffix, 60 IUPAC (International Union of Pure and Applied Chemistry), 51, 1011 Joule (J), 162, 894 Joule, James, 162 Jupiter, 283 Karat scale, 511 Kelvin, Lord (William Thomson), 390 Kelvin scale, 15, 17, 390 Kerosene, 1014 Ketones, 780, 1024–25 nomenclature, 1047 Kidney stones, 722 Kilogram (kg), 15, 16 Kilojoules (kJ), 162 Kilometer, 208 Kilo prefix, 16 Kilowatt-hour (kWh), 849 Kimax, 952 Kinetic energy, 160–62
temperature and, 578 Kinetic-isotope effect, 921 Kinetic-molecular theory, 402–5 gas laws and, 404–5 Kinetics, chemical. See Reaction rates Knocking characteristics of gasoline, 1014 Kroto, Harry, 498 Krypton (Kr), 232, 276 in atmosphere, 751 Krypton compound, 925 Labels, food, 188 Lactic acid, 379, 697, 709, 1025 Lactoferrin, 979 Lactones, 780 Lactose, 1035, 1036 Lanthanide contraction, 965 Lanthanides, 232–33 Lanthanum (La), 155, 233 Large Hadron Collider (LHC), 884–85 Lattice energy, 292–95 calculation of, 295 magnitudes of, 293 Lattice point, 465 Lattice structure, crystal, 469–70 Lattice vectors, 465 Lauryl alcohol, 552 Lauterbur, Paul, 228 Lavoisier, Antoine, 78, 930 Lawrence, Ernest, 911 Laws of combining volumes, 390 of conservation of mass, 78 of constant composition, 10, 40 of definite proportions, 10 of mass action, 614 of multiple proportions, 40–41 scientific, 15 LCDs (liquid crystal displays), 451 LDPE (low-density polyethylene), 494, 495 Lead (Pb), 8, 950 oxidation in aqueous solution, 136 thermodynamic quantities for, 1060 Lead–206, 888, 889 Lead-acid battery, 855 Lead(II) chloride, 623 Lead crystal glass, 952 Lead iodide, 119, 120 Lead nitrate, 119, 120 Lead(II) or plumbous ion (Pb2+), 60 Lead poisoning, 975 Le Châtelier, Henri-Louis, 631, 648 Le Châtelier’s principle, 631–37, 704 catalyst effects, 637–40 change in reactant or product concentrations, 632–33 nitric oxide emission control and, 640 temperature changes, 634–37 volume and pressure changes, 633–34 LEDs (light-emitting diodes), 21, 248–49, 491 Lemon juice, as household acid, 124 Length, SI unit of, 15, 16–17 Leucine, 1031 Leukemia, 904 Leveling effect, 656 Levorotary, 985 Lewis, G. N., 290, 296, 689 Lewis acids and bases, 689–92 crystal-field theory and, 987 metal-ligand bond and, 971
Index Lewis structures, 297–98 alternative, 307–9 dominant, 307–8 drawing, 305–9 with a multiple bond, 306 for a polyatomic ion, 306–7 formal charge and, 307–9 Lewis symbols, 290 LHC (Large Hadron Collider), 884–85 Life, entropy and, 800 LifeStraw, 769–70 Ligand exchange reaction, 1003 Ligands, 968, 974–79 bidentate, 974 color effects of, 985 in living systems, 976–79 monodentate, 974 nomenclature of, 979–80 polydentate (chelating agents), 974–75, 998 weak field and strong field, 989 Ligand-to-metal charge-transfer (LMCT) transition, 993, 1001 Light electron excitation by, 370 monochromatic, 213 scattering by colloidal particles, 541–42 speed of, 208, 209, 214 visible, 208, 209 wave nature of, 208–10 Light absorption, molecular orbitals and, 370 Lightbulbs, energy-efficient, 21 Light-emitting diodes (LEDs), 21, 248–49, 491 Light water reactors, 899 “Like dissolves like,” 522 Lime (calcium oxide), 83, 759, 948 slaked, 199 standard enthalpy of formation for, 184 Lime-soda process, 770 Limestone. See Calcium carbonate (limestone) Limiting reactants (reagents), 99–103 theoretical yields, 102–3 Linear molecular geometry, 333, 334, 335, 337, 340, 343, 351, 1006 Line spectra, 213–14 Line structures, 1013 Linkage isomerism, 981 Lipids, 1037–38 Liquid(s), 7, 437–38 cholesteric, 450 intermolecular attractive forces in, 426 ionic, 436 molecular comparison of solids and, 426–28 nematic, 449, 451 phase changes in, 438–42 properties of, 426, 427 smectic, 449 surface tension of, 437 vapor pressure and, 442–45 viscosity of, 437 volatile, 443 Liquid crystal displays (LCDs), 451 Liquid crystals, 448–52 phases of, 448–49 properties of, 450
Liter (L), 18 Lithium (Li), 8, 49 electron configuration of, 230, 231 formation of, 903 Lewis symbol for, 290 molecular orbitals for, 361, 479 oxidation in aqueous solution, 136 properties of, 269 reaction with oxygen, 270 thermodynamic quantities for, 1060 Lithium cobalt oxide, 258 Lithium drugs, 271 Lithium fluoride, 428 Lithium hydroxide, 99 Lithium ion (Li+), 60 as oxidizing agent, 843 Lithium-ion battery, 258, 856 Lithium–silver chromate batteries, 870–71 Litmus, 127, 663 Living systems. See also Biochemistry chirality in, 1030 iron in, 978–79 metals and chelates in, 976–79 radiation in, 902–8 LMCT transition, 993, 1001 Lobes, in orbitals, 224 Lock-and-key model, 593 Logarithms, 1053–55 antilogarithms, 1054 common, 1053–54 natural, 1054 pH problems using, 1055 London, Fritz, 429–30 London dispersion forces, 429–30, 435 in DNA, 1040 Lone pairs, 334, 676 Lotus plant, 424–26 Low-density polyethylene (LDPE), 494, 495 Lowest unoccupied molecular orbital (LUMO), 370 Lowry, Thomas, 652 Low-spin complex, 991 Lubricants, 1014 Luminous intensity, 15 LUMO (lowest unoccupied molecular orbital), 370 Lung cancer, 906 Lyman series, 242 Lysine, 1031 Lysozyme, 552, 593 Maalox, 130 McMillan, Edwin, 52 Macromolecules, 495 Macroscopic realm, 5 Magic numbers, 882 Magnesite, 948 Magnesium (Mg), 8, 272, 419, 476 combustion of, 82 electron affinity of, 264 electron configuration of, 233 as essential nutrient, 272 Lewis symbol for, 290 oxidation of, 136, 858, 859 properties of, 272 reactions of with acid, 134 with titanium tetrachloride, 867 in seawater, 765 thermodynamic quantities for, 1060–61
Magnesium fluoride, 483 Magnesium hydroxide, 128 Magnesium ion (Mg2+), 60 Magnesium metal, combustion of, 82 Magnesium sulfate, 516 Magnet, permanent, 967 Magnetic moment, 967 Magnetic quantum number, 220 Magnetic resonance imaging (MRI), nuclear spin and, 228 Magnetism, 967–68, 987 antiferromagnetism, 967–68 diamagnetism, 366–68 ferrimagnetism, 968 ferromagnetism, 967 paramagnetism, 366–68, 967, 987 Magnitude, of equilibrium constants, 619–20 Main-group (representative) elements, 233 Malic acid, 651 Malignant tumor, 907 Malleability, 469, 478 Manganese, oxidation in aqueous solution, 136 Manganese dioxide, 930 Manganese(II) or manganous ion (Mn2+), 60 Manganese(II) oxide, 304 Manganese(VII) oxide, 304 Manganese silicide, 961 Manganous or manganese(II) ion (Mn2+), 60 Manhattan Project, 898 Manic-depressive illness, 271 Manometer, 387, 388 Mansfield, Peter, 228 Many-electron atoms, 226–29 Marble, corrosion of, 758 Marconi, Guglielmo, 754 Marsden, Ernest, 43 Marsh, James, 961 Marsh test, 961 Mass(es) atomic, 45. See also Stoichiometry average atomic. See Atomic weights calculating numbers of molecules and atoms from, 92 conservation of, 78 critical, 897 electron, 43 interconverting mole and, 90–91 interconverting number of particles and, 91–92–92 molar, 88–89 calculating, 89 defined, 88 determining through colligative properties, 539–40 effusion rate and, 407–8 gas density and, 395–97 neutralization reaction and, 144 neutron, 45 in nuclear reaction, change in, 894–95 proton, 45 SI unit of, 16–17 subcritical, 897 supercritical, 897 weight vs., 16n Mass action, law of, 614 Mass defect, 895–96 Mass numbers, 46–47, 876
I-13
Mass percentage, 526–27 Mass spectrometer, 49 Mass spectrum, 49 Materials, modern liquid crystals, 448–52 types of liquid crystalline phases, 448–49 for nanotechnology, 496–501 carbon nanotubes, 498–99 metals on the nanoscale, 498 semiconductors on the nanoscale, 497–98 for optics liquid crystals, 448–52 semiconductor light-emitting diodes, 491 polymers and plastics, 490–96 making polymers, 492–94 recycling, 494 structure and physical properties of polymers, 494–96 semiconductors semiconductor doping, 489–90 semiconductor light-emitting diodes, 491 Mathematical operations, 1051–56 exponential notation, 1051–53 logarithms, 1053–55 antilogarithms, 1054 common, 1053–54 natural, 1054 pH problems using, 1055 quadratic equations, 1055–56 Matter classifications of, 7–11 compounds, 8–10 elements, 7–8 mixtures, 7, 8, 10–11 pure substances, 7 conservation of, 40, 78 defined, 4 properties of, 11–14 chemical, 11, 39 extensive, 11 intensive, 11 physical, 11, 39 physical and chemical changes, 12–13 quantitative, 14 separation of mixtures, 13–14 states of, 7, 427 wave behavior of, 216–19 Matter waves, 216–17 Mayan civilization, 115 Meals-ready-to-eat (MREs), 203 Mean free path diffusion and, 408–9 of metals, 498 Measurement, 14–30 dimensional analysis, 25–30 conversion factors in, 25–28 conversions involving volume, 28–29 SI units of, 15–17 base units, 16 density, 19 derived, 18 length and mass, 15, 16–17 for speed, 18 temperature, 17–18 volume, 18–19
I-14
Index
Measurement (cont.) uncertainty in, 20–25 precision and accuracy, 21–22 significant figures, 22–25 uncertainty principle and, 218 Mechanical work, 172 Mechanism, of addition reactions, 1019 Medical imaging, 228 Medicine chelating agents in, 975 chiral drugs in, 1029 radiotracers used in, 893 Medium steels, 474 Mega prefix, 16 Meitner, Lise, 898 Melting curve, 446 Melting points, 445, 446 intermolecular forces and, 428 of metals, 478 normal, 446 of solids, 464 Mendeleev, Dmitri, 250–51, 280 Meniscus, 438 Menthol, 108 Merck Index, 30 Mercuric or mercury(II) ion (Hg2+), 60 Mercuric oxide dry-cell batteries, 871 Mercury (Hg), 8, 51, 418 meniscus, 438 oxidation in aqueous solution, 136 specific heat of, 176 surface tension of, 437 Mercury(I) or mercurous ion (Hg22+), 60 Mercury(I) oxide, 644 Mercury(II) or mercuric ion (Hg2+), 60 Mesitylene, 94 Mesosphere, 750 Metabolism of glucose, 90, 180 nitric oxide and, 369 peroxide ion by-product of, 933 of proteins, 188 Meta isomer, 376 Metal(s), 50, 51, 464, 468–76, 963–1003 active, 135 group trends for, 268–72 activity series of, 135–37 alkali. See Alkali (group 1A) metals alkaline earth. See Alkaline earth (group 2A) metals alloys, 473–76 common, 473 defined, 473 heterogeneous, 475 intermetallic compounds, 475 interstitial, 474, 475 solution, 474 steels, 474 substitutional, 474, 475, 476 close packing in, 470–73 conductivity of, 469, 478, 481 corrosion of, 857–59 iron, 858–59 deficiencies of, 976 energy bands in, 487–88 f-block, 234 group trends for, 268–72 ionic compounds and, 56 ions of, 55 in living systems, 976–79 on nanoscale, 498
noble, 135, 592 oxidation of, 857 by acids and salts, 133–35 periodic properties of, 264–67, 481 physical properties of, 478 qualitative analysis of, 736–39 reactions of with halogens, 276 with hydrogen, 273 with nonmetal, 82 with oxygen, 270, 274 standard reduction potentials and activity series of, 846 structures of, 469–70 transition. See Transition metals Metal cations, acid-dissociation constants for, 682 Metal chlorides, 737 Metal complexes, 968–74 charges, coordination numbers, and geometries, 972–74 metal-ligand bond, 971–72 Werner’s theory of, 969–71 Metal hydroxides, 125 Metal ions acidic solutions and, 683 complex ion formation and, 731–33 coordination numbers of, 969 Lewis acids and bases and, 731–33, 963–64 in water, 682 Metallic bonds, 289, 290, 476–81 electron-sea model, 478 molecular-orbital model, 478–81 Metallic character, 265 Metallic elements, 51. See also Metal(s) Metallic hydrides, 924 Metallic hydrogen, 283 Metallic radius, 280 Metallic solids, 464, 468–76 Metal–ligand bond, 971–72, 987 Metalloenzymes, 594 Metalloids, 51 periodic properties of, 264, 265, 268 Metallurgy, 964 defined, 964 electrometallurgy of aluminum, 862 Metal oxides, 266–67 ionic, 666 Metal-to-ligand charge-transfer (MLCT) transition, 993 meta- prefix, 1021 Metathesis reactions, 121–22 Meter (m), 15, 16, 208 redefinition of, 209 Methanal (formaldehyde), 354, 930, 1006, 1024 in car exhaust gases, 778 Methane, 53, 66, 67, 384, 398, 1006, 1009. See also Natural gas in atmosphere, 751, 764 balanced chemical equation for, 79–80 bond enthalpies in, 315 bonds in, 338, 349, 1009 combustion of, 181, 183, 1008 critical temperature and pressure in, 442 as greenhouse gas, 764 hydrogen production and, 922 molar entropies of, 802
in natural gas, 190 phase diagram of, 447 reactions of with chlorine, 316–17 with iodine, 646 with oxygen, 79 representations of, 54 specific heat of, 176 standard enthalpy of formation for, 184 structural formula for, 54 Methanoic acid. See Formic acid (methanoic acid) Methanol (methyl alcohol), 66, 67, 201, 644, 646, 822, 1022, 1023 combustion of, 1024 dissolution of, 117 hydrogen in manufacture of, 923 reaction with water, 117 solubility of, 521 solution of, 118 standard enthalpy of formation for, 184 Methanol fuel cells, 857 Methionine, 1031 3-Methyl-1-pentene, 1018 2-Methyl-2-propanol, 1023 Methyl acetate, 914 Methylamine, 676, 678, 699, 1003 Methylbenzene (toluene), 108, 486, 532, 551, 773, 1019 Methyl bromide, 574, 757 Methyl chloride, 316–17, 757 Methyl ethanoate, 1022 Methyl ethyl ketone (2-butanone), 1024 Methyl group(s), 1011, 1015 Methylhydrazine, 938, 961 combustion of, 179 Methyl iodide, 460 Methyl isocyanate, 381 Methyl isonitrile, 581 first-order rearrangement of, 573 isomerization of, 577, 600 transformation to acetonitrile, 571, 577–78, 600 Methyl mercaptan, 255 Methyl orange, 664 2-Methylpropane, 1010 Methylpropene, 1015 Methyl propionate, 1026 Methyl red indicator, 664, 721–22 Methyl violet, 664 Metric system, 14 Mexico City, air pollution in, 778 Meyer, Lothar, 250 Mg(OH)2, 724, 728–29 Michelson, A. A., 209 Micrometer, 208 Micro prefix, 16 Microscope, electron, 217 Microstates, 794–96 Microwave radiation, 208, 246 Mild steels, 474 Milk of magnesia, 128 Milk of Magnesia, 130 Milliamp-hours (mAh), 872 Millikan, Robert, 42 Millikan’s oil-drop experiments, 42–43 Milliliter, 18 Millimeter, 208 Millimeter of mercury (mm Hg), 386
Millimolar, 143 Milli prefix, 16 Millivoltmeter, 663 Miscible, 521 Mixing, natural tendency toward, 514 Mixture(s), 7, 8, 10–11 components of, 10 equilibrium, 612–13 of gases, 384 heterogeneous, 10, 11 homogeneous. See Solution(s) racemic, 985, 1029 separation of, 13–14 MLCT transition, 993 Moderator, 898, 899 Modern materials. See Materials, modern Molal boiling-point-elevation constant, 534 Molal freezing-point-depression constant, 535 Molality, 527–29 conversion of, 528–29 Molar entropies, 801–2 Molar heat capacity, 175 Molarity (M), 139–40, 527–30 calculating, 139–40 conversion of, 529–30 interconverting moles, volume, and, 140–41 Molar mass, 88–89 calculating, 89 defined, 88 determination of, 539–40 effusion rate and, 407–8 gas densities and, 395–97 Molar solubility, 723 Mole, 15, 86–92 converting to number of atoms, 87 defined, 86 interconverting masses and, 90–91 numbers of particles and, 91–92 interconverting molarity, volume, and, 140–41 molar mass, 88–89 Molecular architecture, 331 Molecular collisions, 576 Molecular compounds, 53 binary, 65–66 electrolytic behavior of, 126 in water, 118 Molecular diffusion, 405–9 Molecular effusion, 405–9 Molecular equations, 122–23 for redox reactions, 135 Molecular formulas, 53, 54 of complex ion, 972–73 from empirical formula, 94 Molecular geometry, 330–81 bent, 333, 334, 337, 343 covalent bonding and, 345–46 defined, 336 dispersion forces and, 430 hybrid orbitals and, 346 involving d orbitals, 349–50 sp, 346–48 sp2 and sp3, 348–50 intermolecular attraction and, 430 linear, 333, 334, 335, 337, 340, 343, 351 of metal complexes, 973–74 molecular orbitals (MO) and, 358–60 from 2p atomic orbitals, 362–65
Index bond order and, 360 in hydrogen molecule, 358–60 light absorption and, 370 in second-row diatomic molecules, 361–71 molecular (bond) polarity and, 343–45 multiple bonds and, 351–58 octahedral, 333, 335, 339, 340, 341, 343 seesaw, 340 square planar, 334, 340, 341, 343 square pyramidal, 340, 341 tetrahedral, 332, 333, 334, 335, 337, 343, 351 trigonal bipyramidal, 333, 335, 339, 340, 343 trigonal planar, 333, 334, 335, 337, 343, 351 trigonal pyramidal, 333, 334, 336, 337 T-shaped, 333, 340 valance-shell electron-pair repulsion (VSEPR) model of basis of, 335–36 for larger molecules, 342–43 for molecules with expanded valence shells, 339–41 nonbonding electrons and multiple bonds, 338–39 valence-bond theory and, 345, 350 valence-shell electron-pair repulsion (VSEPR) model of, 334–43 Molecular hydrides, 924 Molecularity, 581, 582–83 rate laws for elementary reactions and, 583–84 Molecular orbital diagram. See Energylevel diagram (molecular orbital diagram) Molecular-orbital model for metals, 478–81 Molecular orbitals (MO), 358–60 antibonding, 358–59, 360n from atomic orbital wave functions, 364 bonding, 358 energy and, 370 highest occupied (HOMO), 370 lowest unoccupied (LUMO), 370 molecular geometry and from 2p atomic orbitals, 362–65 bond order and, 360 in hydrogen molecule, 358–60 light absorption and, 370 in second-row diatomic molecules, 361–71 phases in, 363–64 pi (p), 362–65 sigma (s), 359 Molecular orbital theory, 358 of coordination compounds, 994 Molecular oxygen, 757 Molecular perspective, 4–5 Molecular (bond) polarity, 343–45 Molecular solids, 486 Molecular speed, distributions of, 403–4 Molecular weights, 71, 85 boiling point and, 431 Molecule(s), 4, 52–54 aromatic, 311 chemical formulas and, 52–53 of compound, 8 defined, 53
degrees of freedom of, 798, 799, 800, 801–2 diatomic, 53 heteronuclear, 369–71 homonuclear, 361–71 electron density in, 296 of element, 8 optically active, 984–85 organic, 1006–7 picturing, 54 polar, 301, 429–30 properties of states and, 7 Mole fraction, 527–29 conversion of, 528–29 partial pressures and, 400–401 Molina, Mario, 756 Molybdenum, 242 Momentum, 216 Monatomic ions, 132 Monoatomic gases, 384 Monochromatic light, 213 Monochromatic radiation, 213 Monochromatic wavelength, 213 Monoclinic lattice, 466 Monodentate ligands, 974 Monomers, 490 mono- prefix, 65 Monoprotic acids, 124, 664 Monosaccharides, 1035–36 Monosodium glutamate (MSG), 108 Monounsaturated fatty acid, 1037 Montreal Protocol on Substances That Deplete the Ozone Layer, 757 Mortar, 948 Moseley, Henry, 251 Mothballs, 325 Motif, 467 Motion, 160 Motion, molecular, 796–97 Motor oils, 437 Mount Pinatubo, 756 Mount San Antonio, 209 Mount St. Helens, 418 Mount Wilson, 209 MREs (meals-ready-to-eat), 203 MRI (magnetic resonance imaging), 228 MSG (monosodium glutamate), 108 Multiple bonds, 298 bond angles and, 338–39 bond enthalpies of, 316 electron domains for, 339 Lewis structure with, 306 molecular geometry and, 351–58 Multiple-choice questions, 103 Multiple proportions, law of, 40–41 Multiplication in exponential notation, 1052 significant figures in, 24 Multistep mechanisms, 582–83 rate-determining step for, 584–85 Multiwall carbon nanotubes, 499 Murad, Ferid, 941 Mylanta, 130 Myoglobin, 976, 1034 Naming compounds. See Nomenclature Nanomaterials, 496–501 carbon nanotubes, 498–99 metals, 498 semiconductors, 464 Nanometer, 208
Nanoparticles, of platinum, 605 Nano prefix, 16 Naphthalene, 325, 381, 1019 Natural gas, 190, 255, 398, 423, 781, 824. See also Methane combustion of, 781 conversion of, 824 fuel value and composition of, 190 Natural logarithms, 1054 Nebula, 903 Néel temperature, 968 Negative exponent, 1051 Negligible acidity, 656–57 Nematic liquid crystalline phase, 449, 451 Neon (Ne), 49 in atmosphere, 751 atomic emission of, 213 electron configuration of, 230, 231 Lewis symbol for, 290 light emitted by, 206–7, 213 phase diagram for, 458 properties of, 276 Neopentane (2,2-dimethylpropane), 430, 818, 1010 Neo- prefix, 1010 Neptunium, 886 Nernst, Walther, 849 Nernst equation, 849–51 Net ionic equations, 123 for oxidation-reduction reactions, 135 writing, 123–24 Neurotransmitter, nitric oxide as, 940 Neutralization reactions, 127–30 with gas formation, 129–30 using mass relations in, 144 writing chemical equations for, 129 Neutral solutions, 659, 683–85 Neutron diffraction, 243 Neutrons, 44, 45, 876, 878, 879, 885 mass of, 45 Neutron-to-proton ratio, 880–82 Newton (N), 385 Newton, Isaac, 405 Niacin, 669, 671, 673 Nickel (Ni) alloyed with gold, 477 electronic band structure for, 480 electroplating with, 860 reactions of with hydrochloric acid, 517–18 with oxygen, 266 as reducing agent, 846 Nickel(II), 973 Nickel-acid reaction, 517–18 Nickel-cadmium (nicad) battery, 829, 856 Nickel carbonyl, 421 Nickel-metal-hydride battery, 856 Nickel(II) or nickelous ion (Ni2+), 60 Nicotine, 108–9 Nitrate anion, 120 Nitrate ion, 62, 63 delocalized bonding in, 356 resonance structures in, 310 Nitric acid, 124, 125, 664–65, 939–40 reactions of with copper, 12 with gold, 138 Nitric oxide(s), 647, 939 in atmosphere, 751, 760–61 emissions of, 640 energy-level diagram for, 369
I-15
reactions of with bromine, 586–88 with chlorine, 647 with oxygen gas, 798 Nitride ion, 63 Nitrite ions, 563 Nitrobenzene, 1020 Nitrocellulose, 940 Nitrogen (N), 8, 329, 442, 937–41 in atmosphere, 382, 384, 751 deep-sea diving and, 525 dissociation energy of, 754 distribution of molecular speeds for, 403 electron configuration of, 230–31 fixed, 937 hydrogen compounds of, 937–39 Lewis symbol for, 290 melting and boiling points of, 428 molecular, 367 bonding in, 298 critical temperature and pressure in, 442 ionization of, 754 Lewis structure of, 298 photodissociation of, 754 properties of, 751 specific heat of, 176 mole relationships of, 88 oxidation states of, 937 oxides and oxyacids of, 939–41 production and uses of, 937 properties of, 937, 941, 942 reduction of, 594 2p orbital filling in, 261 Nitrogen–13, 893 Nitrogen–14, 887 Nitrogenase, 594 Nitrogen cycle, 594 Nitrogen dioxide, 12, 384, 616, 782, 798, 939, 940 decomposition of, 572 dinitrogen tetroxide-nitrogen dioxide equilibrium, 616, 621–22 photodissociation of, 779 in smog, 760 Nitrogen fixation, 594, 638 Nitrogen gas, in air bags, 398–99 Nitrogen oxides, 592, 758, 760–61 high-temperature behavior of, 603 Nitroglycerin, 193, 319, 940, 941 Nitro isomer, 981 Nitrosyl bromide, decomposition of, 647 Nitroto isomer, 981 Nitrous acid, 940 properties of, 667 reaction with water, 655 Nitrous oxide (laughing gas), 384, 939 in atmosphere, 751 decomposition of, 586 NMR (nuclear magnetic resonance), 228 Nobel, Alfred, 193, 319, 941 Noble-gas compounds, 925 Noble-gas core, 231 Noble (rare) gases (group 8A), 51, 924–25 boiling points of, 429 electron affinities for, 264 group trends for, 276–77 Noble metals, 135 Nodal plane, 362, 364
I-16
Index
Nodes, 222 on waves, 219 Nomenclature, 59 of alkanes, 1010–13 of alkenes, 1015 of alkynes, 1017 amino acids, 1031 chemical, 59 of coordination compounds, 979–80 of inorganic compounds, 59–66 acids, 64–65 binary molecular compounds, 65–66 ionic compounds, 59–64 Nomex, 509 Nonane, 1009 viscosity of, 437 nona- prefix, 65 Nonbonding atomic radius, 254 Nonbonding electrons, bond angles and, 338–39 Nonbonding pair, 334, 338–39 Nonelectrolyte, 117 Nonionizing radiation, 902 Nonmetallic elements, 50, 51 Nonmetals, 50, 51, 916–61 boron, 953–54 electron configuration of, 233 Lewis symbol for, 290 electrons added to, 294 group 4A, 949–52. See also Carbon (C); Silicon (Si) general characteristics of, 949–50 group 5A, 941–45. See also Nitrogen (N); Phosphorus (P) general characteristics of, 941–42 group 6A, 934–36. See also Oxygen (O) general characteristics of, 934 group trends for, 273–74 occurrences and production of, 934 oxides, oxyacids, and oxyanions of sulfur, 935–36 properties and uses of, 934–35 sulfides, 935 group 7A (halogens), group trends for, 274–76 group 8A (noble gases), 924–25 compounds, 925 group trends for, 276–77 hydrogen. See Hydrogen (H) ions of, 55 oxidation number of, 132 oxygen group (6A), group trends for, 273–74 periodic properties of, 264–65, 267–68, 918–20 reactions of, 919–20 with hydrogen, 273 with metals, 82 Nonpolar covalent bond, 299 Nonpolar molecules, 343–44 Nonrenewable energy, 191 Nonspontaneous reactions, 786–87, 814 Nonvolatile substance, 530–31 Normal boiling point, 444, 811–12 Normal melting point, 446 Novoselov, Konstantin, 500 n-type semiconductor, 490 Nuclear age, 898 Nuclear binding energies, 895–96
Nuclear charge effective, 251–54 estimating, 253 for period 2 and period 3 elements, 253 for valence electrons, 252, 254 ionic radii and, 256 Nuclear chemistry, 52, 874–915 biological effects of radiation, 900, 902, 904 dosage and, 904–5 radon, 906 therapeutic, 875, 893, 907 cancer treatment with, 875, 907 defined, 875 elementary particles in, 876 energy changes in nuclear reactions, 894–96 nuclear binding energy, 895–96 fission, 896–901 reactors using, 898–900 waste from, 900–901 fusion, 896, 902 nuclear stability patterns even vs. odd number of nucleons, 882, 883 magic numbers and, 882 neutron-to-proton ratio, 880–83 radioactive series (nuclear disintegration series), 882 nuclear transmutations, 884–86 radioactive decay, 877, 878–80 rates of, 886–91 types of, 878–80 radioactivity, 876–80 detection of, 891–92 Nuclear disintegration series, 882 Nuclear energy, 191 Nuclear equations, 877–78 balanced, 884 writing, 879–80, 884 Nuclear fission reactor, 898–900 Nuclear force, 46, 880 Nuclear magnetic resonance (NMR), 228 Nuclear model of the atom, 43–44 Nuclear power plant design, 899–900 Nuclear reactions. See Nuclear chemistry Nuclear reactors, 898–900 Nuclear spin, magnetic resonance imaging and, 228 Nuclear transmutations, 884–86 Nuclear waste, 900–901 Nucleic acids, 1038–42 Nucleons, 876 Nucleotides, 1040 Nucleus (atomic), 44 shell model of, 882 Nuclide, 876 Numbers exact vs. inexact, 20–21 rounding off, 24 Nutrition label, 188 Nylons, 492, 493, 494, 949 Oblique lattice, 465, 466 Obsidian (volcanic glass), 465 Oceans, 512–13, 765–66 acidification of, 703–4, 728 temperature of, climate change and, 781
Octahedral complexes, electron configurations in, 990–91 Octahedral crystal field, 987–88 Octahedral geometry, 333, 335, 339, 340, 341, 343, 973 Octane, 66, 67, 200, 201, 437, 821, 1009 solution formation and, 517 viscosity of, 437 Octane numbers, 1014 Octane rating, 1014 octa- prefix, 65 Octet, 231 Octet rule, 290–91 exceptions to, 312–14 n-Octylglucoside, 549 Odors, 680, 699 esters and, 1026 Ogallala aquifer, 779 OH group, acid-base equilibria and, 686–87 Oil combustion of, 758–59 crude (petroleum), 190, 1014 Oil-drop experiments, Millikan’s, 42–43 Olefins, 1008 Oleic acid, 1037 Omega-3 and omega-6 fatty acids, 1038 Omega Nebula, 2 Open system, 162 Opsin, 357 Optical isomers (enantiomers), 983–84, 1028–29, 1030 Optically active molecules, 984–85 Orbital diagram, 229 electron configurations and, 231 Orbital overlap, 345–46 Orbitals atomic. See Atomic orbitals hybrid, molecular geometry, and, 346–51 molecular. See Molecular orbitals (MO) valence, 249 Ores, 964 Organic chemistry, 66, 1004–50 chirality in, 1028–29 compounds with carbonyl group aldehydes and ketones, 1024–25, 1047 amine and amides, 1028 carboxylic acids, 1025–27 esters, 1025–27 functional groups, 1007, 1021–28 alcohols, 1023–24 aldehydes and ketones, 1024–25 amine and amides, 1028 carboxylic acids and esters, 1025–27 ethers, 1024 general characteristics of organic molecules, 1006–7 hydrocarbons, 66–67, 1008–14 alkenes, 1015–17 alkynes, 1017–19 aromatic, 1008, 1019–21 branched-chain, 1009 saturated (alkanes), 66–67, 1008, 1009–14 straight-chain, 1009 Organic compounds, 59, 66–67 volatile, 773
Organic molecules, structures of, 1006–7 Organic substances, stability of, 1007 Orientation factor in reaction rates, 576, 577 Orlon, 949 Ortho isomer, 376 ortho- prefix, 1021 Orthorhombic lattice, 466 Osmosis, 536–39 in living systems, 538–39 reverse, 768–69 Osmotic pressure, 536–37, 538 molar mass from, 539–40 Ostwald process, 939 -ous suffix, 60, 64 Outer-shell electrons, 231–32 Overall reaction order, 565 Overhydration, 143 Overlap, orbital, 345–46 Oxalate ion, 994 Oxalic acid, 651, 652 acid-dissociation constant of, 674 Oxidation, 131 of alcohols, 1024–25 of glucose, 814 of iron, 131, 787 of metals, 133–35 Oxidation numbers (oxidation states), 132–33, 265–66, 828–29 acidity and, 687 formal charge and, 309 of transition metals, 965–67 Oxidation potential, 873 Oxidation reactions, 84 Oxidation-reduction equations, 830–35 Oxidation-reduction (redox) reactions, 131–38, 828–35 activity series and, 135–37 balancing, 830–35 in basic solution, 833–35 half-reaction method, 830–33 in batteries, 855 corrosion, 857–59 of iron, 858–59 defined, 131 disproportionation, 872, 940 electron movement in, 828–29, 830, 831, 832, 837–38 free energy and, 845–49 molecular and net ionic equations for, 135 of nitrogen, 594 oxidation numbers (oxidation states), 132–33 oxidation of metals by acids and salts, 133–35 spontaneity of, 835, 845–46, 847 in voltaic cells, 835–37 concentration cells, 852–54 emf in, 838–45 Oxide(s), 931, 932–33 acidic, 932 amphoteric, 733–34 basic, 932–33 of boron, 953 of carbon, 946–48 of nitrogen, 939–41 sulfur, 935–36 Oxide ion, 63, 274 Oxidizing agent (oxidant), 829 strengths of, 843–45
Index Oxyacetylene, 1017 Oxyacids, 686–88, 929 of halogens, 929 of nitrogen, 939–41 sulfur, 935–36 Oxyanions, 61, 62, 929 common, 61 of halogens, 929 as oxidizing agents, 843 of sulfur, 935–36 Oxy compounds of phosphorus, 942–44 Oxygen (O), 8, 442, 925, 930–33 allotropes of, 273. See also Ozone in atmosphere, 382, 384, 751 in blood, 713 dissociation energy of, 753–54 electron configuration of, 249 as excess reactant in combustion reactions, 100 formation of, 903 in green chemistry, 773 Lewis symbol for, 290 methane reacting with, 79 molecular, 4, 9, 53 bonding in, 367–68 combustion reactions with, 919 critical temperature and pressure in, 442 ionization of, 754 Lewis structure for, 367 paramagnetism of, 368 photodissociation of, 753–54 properties of, 751 oxidation number of, 132 oxides of, 931, 932–33 as oxidizing agent, 843, 930 ozone, 930 paramagnetism of, 368 peroxides, 933 production of, 930–31 properties of, 9, 273, 930 reactions of with alkali metals, 270 dissolution in water, 768 with hydrogen, 173 with metals, 270, 274 with methane, 79 with nickel, 266 with nitric oxide, 798 with sulfur tetrafluoride, 380 solubility of, 526 superoxides, 933 2p orbital filling in, 261 uses of, 931 Oxygen anions, 274 Oxygen atom, 9 Oxygen-demanding wastes, 768 Oxygen group. See Group 6A elements (chalcogens) Oxyhemoglobin, 976–77, 1001 Oxymyoglobin, 977 Ozone, 53, 199, 273–74, 379, 930, 931–32, 1024 in atmosphere, 751, 754–56, 758, 932 concentration in air, 112 decomposition of, 646 molecular structure for, 309 reaction with chlorine, 756–57 resonance structures in, 309–10 in smog, 761 structure of, 931
water disinfection with, 771 Ozone hole, 756 Ozone layer, 246, 755–57 depletion of, 756–57, 782 halogen-containing molecules and, 574 photodecomposition, 756–57 Pacemaker cells, 853 Packing efficiency, 472–73 Paired electron, 229 Palladium, 924 alloyed with gold, 477 Palladium(II), 970 Paper chromatography, 14 Paraffins, 1014 Para isomer, 376 Parallelepipeds, 466 Parallel spins, 230 Paramagnetism, 366–68, 967, 987 para- prefix, 1021 Partial charges, 309 Partial pressures, 399–401 mole fractions and, 400–401 pressure-volume changes and, 634 Particle accelerators, 884–85 Parts per billion (ppb), 526–27 Parts per million (ppm), 526–27, 751 Pascal (Pa), 385, 801n Pascal, Blaise, 385 Paschen series, 245 Pattern recognition, 58 Pauli, Wolfgang, 227 Pauli exclusion principle, 227–29 Pauling, Linus, 299 p-block elements, electron configurations of ions of, 294 Pearlite, 475 “Pebble-bed” reactor design, 900 p elements, electron affinities for, 264 Pentaborane, 203 Pentane, 1009, 1010 n-Pentane, 430, 818 Pentanol, 521 penta- prefix, 65 Pentene, isomers of, 1016–17 Pentyl acetate, 1026 Peptide bonds, 1030–32 Pepto-Bismol, 276 Percentage composition, 85–86 empirical formula from, 93 Percent ionization, 669, 671–73 acid-dissociation constant to calculate, 673 concentration and, 672 Percent yield, 102–3 Perchlorate, 929–30 Perchlorate ion, 62, 63 Perchloric acid, 125, 687, 929 Period 2 diatomic molecules, 361–71 Periodicity, 208 Periodic properties of elements, 248–87 atomic radii, 254–59 periodic trends in, 255–56 effective nuclear charge (Zeff), 251–54 electron affinities, 263–64 group trends for active metals, 268–72 alkali metals (group 1A), 268–72 alkaline earth metals (group 2A), 272 group trends for nonmetals, 273–77
halogens (group 7A), 274–76 hydrogen, 273 noble gases (group 8A), 276–77 oxygen group (group 6A), 273–74 ionic radii, 254–59 periodic trends in, 256–59 ionization energy, 259–63 electron configurations of ions and, 262–63 periodic trends in, 260–62 variation in successive, 259–60 metalloids, 268 metals, 264–67 nonmetals, 267–68 Periodic table, 8, 49–52 development of, 250–51 electron configurations and, 233–38 groups in, 50, 51 ionic charges and, 56 metallic elements or metals, 50, 51 metalloids, 51 nonmetallic elements or nonmetals, 50, 51 periods of, 51 transuranium elements, 52 Periodic trends, 918–20 in atomic radii, 255–56 in first ionization energies, 260–62 in ionic radii, 256–59 Periods, 50, 51 Permanent magnet, 967 Permanganate half-reaction, 831 Permanganate ion, 63, 993 Peroxidase, 933 Peroxide ion, 63, 274, 378, 933 Peroxides, 270, 933 per- prefix, 61, 64 Perspective drawing, 54 PES (photoelectron spectroscopy), 285, 286 PET (polyethylene terephthalate), 493, 773 PET (positron emission tomography), 875, 893 Peta prefix, 16 Petroleum, 190, 1014 global population growth and demand for, 192 PF5, 313 pH, 660–64. See also Acid-base equilibria; Aqueous equilibria of buffer, 708–10 calculating acid-dissociation constant from, 668–69 calculating from acid-dissociation constant, 670–73 of common substances, 663 determining, using concentration cell, 854 logarithms to solve problems of, 1055 measuring, 663–64 salt effects on, 681–85 of seawater, 728 solubility and, 728–30 of strong acid, 665 of strong base, 665–66 titration curve, 714 Phase changes in liquids, 438–42 critical temperature and pressure and, 441–42 energy changes accompanying, 439–40 entropy change and, 791 heating curves, 440–41
I-17
Phase diagrams, 445–48, 534 Phases in atomic and molecular orbitals, 363–64 condensed, 426 ortho-Phenanthroline, 994 Phenol, 1023 properties of, 486 Phenol acid, 667, 668 Phenolphthalein, 145, 146, 664, 721, 722 Phenylacetic acid, 697 Phenylalanine, 1031, 1032 Phenylamine, 1028 Phenylmethanamide, 1028 Phenyl methanoic acid (benzoic acid), 179, 667, 688, 697, 1025 properties of, 667 Pheromones, 1004, 1005 pH meter, 663, 853 Phosgene, 320–21, 339, 619 Phosphate ion, 62, 63 Phosphate ion, Lewis structures for, 314 Phosphates insoluble, 738 as sequestering agents, 975 Phosphine, 442 Phospholipids, 1038–39 Phosphoric acid, 928, 943, 944 acid-dissociation constant of, 674 sale of, 6 Phosphorous acid, 691–92, 943 Phosphors, 892 Phosphorus (P), 8, 476 allotropes of, 942 elemental, 329 halides of, 942 Lewis symbol for, 290 nonbonding electron pairs in, 477 occurrence, isolation, and properties of, 942 oxy compounds of, 942–44 properties of, 941, 942 red, 942 silicon doping with, 489–90 white, 942 Phosphorus–32, 893 Phosphorus halides, 942 Phosphorus(III) oxide, 942 Phosphorus(V) oxide, 943 Phosphorus pentachloride, 644 Phosphorus trichloride, 942 Phosphorus trihalides, 380 Photocell, 245 Photochemical smog, 592, 760–61 Photocopiers, 935 Photodissociation, 752, 753, 756–57 Photoelectric effect, 210, 211–12 Photoelectron spectroscopy (PES), 285, 286 Photographic plates/film, 891 Photoionization, 752, 754 Photoluminescence, 497–98 Photons, 210–12 Photoreceptors, 357 Photosynthesis, 193, 978 Photovoltaics, 5, 193 pH range, 711 pH titration curve, 714, 716, 717 Physical changes, 12–13 Physical properties, 11, 39 Phytoplankton, 728
I-18
Index
pi (p) bonds, 352–58 in alkenes, 1016 in aromatic hydrocarbons, 1020 in chemistry of vision, 357 delocalized, 356 in double bonds, 352 in ozone, 931 periodic trends and, 918 strength of, 352 in triple bonds, 354–55 Pico prefix, 16 pi (p) molecular orbitals, 362–65 in aromatic hydrocarbons, 1020 Pipelines, gas, 398 Pipets, 18, 19 Pipevine swallowtail caterpillar, 196 Planck, Max, 210–11, 213 Planck’s constant, 211, 214, 218 Plastics, 492 polycarbonate, 493 recycling, 494 thermoset, 492 Platinum, 136, 605 Platinum(II), 973 Plato, 40 Plumber solder, 473 Plumbous or lead(II) ion (Pb2+), 60 Plum-pudding model of the atom, 43 Plutonium (Pu), 52, 233, 886 Plutonium–239, 896, 900–901 p-n diode, 491 pOH, 662–63 Polar covalent bond, 300 Polarity acidity for binary acids and, 685 bonding, 298–304 molecular (bond), 343–45 proton transfer reactions and, 653 solubility and, 520 Polarizability, 429–30 Polar liquids, solubility of, 520 Polar molecules, 301, 343–44, 429–30 Pollutants atmospheric, 758 CFCs, 574, 756–57, 764 hydrocarbons, 761 nitrogen oxides, 760–61 sulfur dioxide, 758–60 in urban atmosphere, 758 Pollution air, 592 real-time analysis for prevention of, 772 smog, 760–61 thermal, 526 water, 944 Polonium (Po), 43, 273, 934 Polonium–218, 906 Polyacetylene, 500–501 Polyacrylonitrile, 509 Polyatomic ions, 55, 61, 306–7 Polycarbonate, 493 Polychloroprene, 509 Polychromatic radiation, 213 Polydentate ligands (chelating agents), 974–75, 998 Polyesters, 492 Polyethylene, 67, 492, 493, 511 high-density, 494, 495 low-density, 494, 495 properties of, 494–95
Polyethylene terephthalate (PET), 493, 773 Polymer(s), 464, 490–96 biopolymers, 1029 co–, 494 commercially important, 493 conducting, 500–501 cross-linking of, 495–96 elastomeric, 492, 825 making, 492–94 structure and physical properties of, 494–96 types of, 492 Polymeric solids, 490–96 Polymerization, 490 addition, 492–93 condensation, 493–94 Polynucleotide, 1040 Polypeptides, 1030–32 Polyphosphate, 497 Polypropylene, 493, 511 Polyprotic acids, 674–75 acid-dissociation constants of, 674 titrations of, 720–21 Polysaccharides, 1036–37 Polystyrene, 493 Polytetrafluoroethylene (Teflon), 510, 773, 927 Polyunsaturated fatty acids, 1037 Polyurethane, 493, 783 Poly(vinyl alcohol), 342 Polyvinyl chloride (PVC), 493, 494, 927 Population, global growth of, 192 p orbitals, 224–25 energy-level diagrams/electron configurations, 365–66 periodic trends and, 918–20 phases in, 363–64 radial probability functions of, 252 Porphine molecule, 976 Porphyrins, 976 Positive exponent, 1051 Positron, 878, 879 Positron emission, 878 Positron emission tomography (PET), 875, 893 Potash (potassium carbonate), 1026–27 Potassium (K), 8, 49, 232 oxidation in aqueous solution, 136 properties of, 269 reaction with oxygen, 270 in seawater, 765 thermodynamic quantities for, 1061 Potassium–40, 904, 907–8, 912 Potassium carbonate (potash), 1026–27 Potassium chlorate, 401, 929, 930 Potassium dichromate, 1024 Potassium iodide, 119, 120 Potassium ion (K+), 60 Potassium nitrate, 119 Potassium superoxide, 287, 933 Potential difference, 838 Potential energy, 160–62 electrostatic, 160–61 free energy and, 804 Powers, in exponential notation, 1052 ppb (parts per billion), 526–27 ppm (parts per million), 526–27, 751 Practice, importance of, 29 Praseodymium, 233 Precipitate, 119 Precipitation, 119–24
exchange (metathesis) reactions, 121–22 of ionic compounds, 722 solubility guidelines for, 120–21 ionic equations, 122–24 of ions, 734–36 selective, 735–36 Precision, 21–22 Prefixes binary compound, 65 Greek, 65, 980 metric system, 16 Pressure, 169, 385–87 atmospheric, 169n, 172, 385–87 blood, 388 critical, 441–42 defined, 385 equilibria and, 633–34 equilibrium constants in terms of, 617–18 gas, 385–87 intermolecular forces and, 427 interrelation between temperature and, 1056 Le Châtelier’s principles and, 631 osmotic, 536–37, 538 partial, 399–401 mole fractions and, 400–401 real vs. ideal gas behavior and, 409–11 solubility and, 523–25 spontaneous processes and, 786–87 standard, 801n vapor, 442–45 Pressure-volume relationship, 388–89 Pressure-volume (P-V) work, 170–71, 172 Pressurized water reactor, 899 Priestley, Joseph, 930 Primary cells (batteries), 855 Primary coolant, 898, 899 Primary structure, of proteins, 1032, 1033 Primary valence, 969 Primitive cubic unit cell, 467, 469, 470, 482 Primitive lattice, 467 Principal quantum number (n), 214, 220 Probability, entropy and, 795 Probability density, 220, 224, 363 Problem solving, 86 procedure, 144 Products, 78 calculating amounts of, 98–99 from limiting reactant, 100–101 change in concentration of, 632–33 enthalpy change and state of, 173, 174 states of, 81 Proline, 1031 Propane, 66, 67, 384, 421, 442, 1008, 1009 carbon-carbon backbone of, 1007 combustion of, 84, 185–86, 807 critical temperature and pressure of, 442 molar entropies of, 802 in natural gas, 190 properties of, 431 rotation about carbon-carbon single bonds of, 1009 standard enthalpy of formation for, 184 states of, 428 1,2,3-Propanetriol (glycerol), 1023, 1037, 1038 1-Propanol, 66–67
Propanol, 521, 820 1-Propanol, hydrogen bonding in, 435 2-Propanol (isopropyl alcohol), 67, 95–96, 1023 Propanone (acetone), 150, 520, 554, 555, 1022, 1024, 1025 Propene (propylene), 6, 377, 1014, 1015 Propenoic acid, 109 Property, 4 Propionic acid, 697 Propyl alcohol, 455 Propylene, 6, 377, 1015 Propyl group, 1011 Propyne, 343 Protein(s), 188, 509, 1029–34 amino acids, 1029–32 amphiprotic behavior of, 689 side chain of, 1032 carbon group in, 58 defined, 1029 DNA structure and synthesis of, 1040–41 as enzymes, 592 fibrous, 1034 globular, 976, 1034 metabolism of, 188 polypeptides and, 1030–32 structure of, 1032–34 Protein sequence, 1049 Protium, 920–21 Proton(s), 44–45, 876, 878, 879 mass of, 45 neutron-to-proton ratio, 880–82 Proton donors, 124 Proton-transfer reactions, 653–54 Proust, Joseph Louis, 10 “p” scales, 662–63 p-type semiconductor, 490 Pure substances, 7, 11 crystalline, 799, 800 Purines, 1049 Putrescine, 680 PVC (polyvinyl chloride), 493, 494, 927 Pyrene, 1019 Pyrex, 952 Pyridine, 676, 998 Pyrimidines, 1049 Pyrite (fool’s gold), 465, 935 Pyrosulfuric acid, 935 Pyruvic acid, 1041–42 Quadratic equations, 1055–56 Qualitative analysis for metallic elements, 736–39 Quantitative analysis, 736 Quantitative information, from balanced equations, 96–99 Quantitative properties, 14 Quantity-volume relationship, 390–91 Quantized energy, 210–12 Quantum, 210 Quantum dots, 111, 497–98 Quantum mechanics, 207, 219–20 Quantum number angular momentum, 220 magnetic (m), 220 orbitals and, 220–22 principal (n), 214, 220 spin magnetic (ms), 227 Quantum theory, 207, 213 Quantum wells, 498
Index Quantum wires, 498 Quartz, 465, 486, 951, 952 Quartz glass (silica glass), 952 Quaternary structure, of protein, 1033, 1034, 1048 Questions in tests, types of, 103 Quicklime (calcium oxide), 83, 759, 948 standard enthalpy of formation for, 184 Quinine, 680, 1049 Racemic mixtures, 985, 1029 rad (radiation absorbed dose), 904 Radial probability density, 222 Radial probability function, 222, 225, 252 Radiant energy, 208 Radiation, 180 alpha (a), 43, 877, 878, 879, 902, 903, 904 background, 905 beta (b), 43, 878, 879, 902, 904 biological effects of, 900, 902, 904 dosage and, 904–5 radon, 906 therapeutic, 875, 893, 907 blackbody, 210 electromagnetic, 208 gamma (g), 43, 208, 209, 878–80, 902, 904, 907 infrared, 208, 209 ionizing, 902 microwave, 208, 246 monochromatic, 213 nonionizing, 902 polychromatic, 213 Radiation therapy, 875, 893, 907 Radicals, free, 904 Radioactive decay, 877, 878–80 rates of, 886–91 types of, 878–80 Radioactive series (nuclear disintegration series), 882 Radioactivity, 43, 876–80 detection of, 891–92 Radiocarbon dating, 887–88 Radioisotopes, 876, 906, 907 Radiometric dating, 887–88 Radionuclides, 876 Radiotracers, 892–93 medical applications of, 893 Radio waves, 208, 754 Radium (Ra), 43 electron configuration of, 233 Radium–226, 877, 912 Radius. See Atomic radius/radii; Ionic radii Radon, 233, 276, 553, 906, 924 Radon–222, 906 Radon-testing kits, 906 Rainwater, 267–68, 758 Randomness. See Entropy(ies) Raoult’s law, 530–31 Rare earth elements, 232–33 Rate constants, 565, 568 temperature and, 576 units of, 567 Rate-determining (rate-limiting) step, 584–85 Rate laws, 563–69 concentration and, 563–69 differential, 569 for elementary steps, 583–84
exponents in, 565–67 H+ concentration and, 661 initial rates to determine, 568–69 integrated, 569–70, 571 for multistep mechanisms, 584–85 units of rate constant, 567 RBE (relative biological effectiveness), 905 RDX (cyclotrimethylenetrinitramine), 328 Reactants (reagents), 78 calculating amounts of, 98–99 change in concentration of, 632–33 enthalpy change and state of, 173, 174 environmentally friendly, 773–75 excess, 100 limiting (limiting reagents), 99–103 theoretical yields, 102–3 physical state of, 558 states of, 81 Reaction(s), 12, 78, 557 acid-base, 124–30. See also Acid-base equilibria electrolytes, 126–27 with gas formation, 129–30 gas-phase, 654 neutralization reactions and salts, 127–29 addition of alkenes and alkynes, 1017–19 mechanism of, 1019 of alkanes, 1013–14 anaerobic, 192 analyzing, 138 aqueous. See Aqueous equilibria; Aqueous reactions; Aqueous solution(s) bimolecular, 581, 583 carbonylation, 1026 chain, 897 chemiluminescent, 575 click, 775 combustion, 76, 83–84 balanced equations for, 84 with oxygen, 919 condensation, 943, 1024 with alcohol, 1026 decarbonylation, 822 displacement, 133–34 disproportionate, 940 elementary, 581, 583–84 endothermic, 167, 168, 171 enthalpies of, 172–74, 316–18 entropy changes in, 798, 800–803 exothermic, 167, 168, 171 first-order, 569–71, 574 Friedel-Crafts, 1021 gas volumes in, 397–99 half-life of, 573–75 heat of, 172–74 involving nonmetals, 918–20 ligand exchange, 1003 mechanisms of. See Reaction mechanisms nonspontaneous, 814 nuclear. See Nuclear chemistry periodic trends and, 918–20 proton-transfer, 653–54 rates of. See Reaction rates redox. See Oxidation-reduction (redox) reactions second-order, 571–72, 575 solution formation and, 517–18
spontaneity of. See Spontaneous processes substitution, 1020–21 termolecular, 581 thermite, 168, 201 thermonuclear, 902 unimolecular, 581 water and, 115 Reaction mechanisms, 557, 581–88 defined, 581 elementary reactions, 581 elementary steps rate laws for, 583–84 with fast initial step, 586–88 multistep, 582–83 rate-determining step for, 584–85 with slow initial step, 585–86 Reaction orders, 565–67 overall, 565 Reaction quotient (Q), 734 Reaction rates, 558–609 average, 560 catalysis and, 559, 589–95 enzymes, 591–95 heterogeneous, 590–91 homogeneous, 589–90 concentration and, 558, 559 change with time, 569–75 rate laws, 563–69 defined, 557, 559 factors affecting, 558–59 instantaneous (initial rate), 561–62 spectroscopic methods to measure, 564 stoichiometry and, 562–63 temperature and, 558, 575–81 activation energy, 577–78, 579–80 Arrhenius equation for, 578–79 collision model of, 576 orientation factor in, 576, 577 time and, 559, 561, 569–75 for weak and strong acids, 672 Reactivity, patterns of, 81–84 combinations and decomposition reactions, 81–83 combustion reactions, 83–84 Reactors, nuclear, 898–900 Reagents. See Reactants (reagents) Real gases, 409–13 van der Waals equation, 411–13 Rectangular lattice, 465, 466 Recycling symbols, 494 Red blood cells, 713 osmosis and, 537–38 sickled and normal, 545 Red giant phase, 903 Red ochre, 993 Redox reactions, 828–35. See also Oxidation-reduction (redox) reactions Red phosphorus, 942 Reducing agent (reductant), 829 strengths of, 844 Reduction. See Oxidation-reduction (redox) reactions Refining, 1014 Reforming, 1014 Refrigerant, 823 carbon dioxide as, 947 Reinecke’s salt, 979 Reinitzer, Frederick, 448 Relative biological effectiveness (RBE), 905
I-19
Relativistic Heavy Ion Collider (RHIC), 884–85 rem (roentgen equivalent for man), 905 Remsen, Ira, 12–13 Renewable energy, 20–21, 191 Renewable feedstocks, 772 Representative (main-group) elements, 233 Resonance structures, 309–12, 355–56 in benzene, 311–12 in nitrate ion, 310 in ozone, 309–10 Retinal, 357 Reverse osmosis, 768–69 Reversible process, 788–90 R groups, 689, 1030, 1031 Rhenium oxide, 504 RHIC (Relativistic Heavy Ion Collider), 884–85 Rhodopsin, 357 Rhombic sulfur, 934 Rhombohedral lattice, 466 Rhubarb, 650, 651 Ribonuclease A, 1050 Ribonucleic acid (RNA), 1038 Ribose, 1040 Ring structure, of glucose, 1035 rms speed, 404, 405–6 RNA (ribonucleic acid), 1038 Rocket fuel, 922, 930, 938, 961 Rods, 357 Rolaids, 130 Rømer, Ole, 209 Roosevelt, Franklin D., 898 Root-mean-square (rms) speed, 404, 405–6 Roots, in exponential notation, 1052 Rotational motion, 796 Rounding off numbers, 24 Rowland, F. Sherwood, 756 Royal Institution of Great Britain’s Faraday Museum, 498 Rubber, 465, 492 vulcanization of, 496, 934 Rubidium, 232, 269 thermodynamic quantities for, 1061 Rubidium–87, 904 Rusting. See Corrosion Rutherford, Daniel, 937 Rutherford, Ernest, 43–44, 251, 884 Rutile, 507, 508 Rutile structure, 483 Rydberg constant, 213–14 Rydberg equation, 213, 215 Saccharin, 697 Sacrificial anode, 859 SAE (Society of Automotive Engineers), 437 Salinity, of seawater, 765–66 Salt(s). See also Sodium chloride acid, 680 chlorate, 929 defined, 128 density of, 19 dissolving of, 798 electrolysis of molten, 860 formula weight of, 88 hypochlorite, 929 iodized, 928 molar mass of, 90 naming, 979 neutralization reactions and, 127–29
I-20
Index
Salt(s) (cont.) oxidation of metals by, 133–35 solubility-pH relationship in, 729–30, 733 Salt bridge, 836–37, 852, 857 Saltpeter, 937 Salt solutions acid-base properties of, 681–85 anion reaction with water, 681–82 cation reaction with water, 682–83 combined cation-anion effect, 683–85 conductivity of, 116, 117 Saponification, 1026 Saturated hydrocarbons. See Alkanes Saturated solution, 518–20, 722–23 Saturn, 283 s-block elements, electron configurations of ions of, 294 Scale, 770 Scandium, 232, 1061 Scandium fluoride, 483 Scanning tunneling microscopy (STM), 41 Schrödinger, Erwin, 219 Schrödinger’s wave equation, 219 Scientific law, 15 Scientific method, 15 Scientific notation, 1051 Scintillation counter, 892 Screen, 252 Screening constant (S), 252 Seaborg, Glenn, 52 Seaborgium, 52 Sea urchins, 728 Seawater, 728, 765–66 desalination of, 768–69 ionic constituents of, 765 Second (s or sec), 15 Secondary cells (batteries), 855 Secondary coolant, 899 Secondary structure, of proteins, 1032, 1033 Secondary valence, 969 Second ionization energy, 259 Second law of thermodynamics, 792–93 Second order overall, 565 Second-order reactions, 571–72, 575 Seesaw geometry, 340 Selective precipitation of ions, 735–36 s elements, electron affinities for, 264 Selenium (Se), 934–35 electron configuration of, 234 properties of, 273 thermodynamic quantities for, 1061 Semiconductors, 268, 487–90 band structure of, 488 compound, 488 doping of, 489–90 elemental, 488 light-emitting diodes, 491 on nanoscale, 497–98 n-type, 490 p-type, 490 silicon in, 950 Semimetals, 500 Semipermeable membranes, 536, 537, 544, 853 Separation, of ions, 734–36 Sequestering agents, 975 Serine, 1031 Serotonin, 112
Serpentine asbestos, 951 Seven Up, 271 Shape, molecular. See Molecular geometry SHE (standard hydrogen electrode), 839–40 Shell model of nucleus, 882 Shiver, 180 Shock sensitive, 319 Sickle-cell anemia, 545, 1032 Side chain, amino acid, 1009, 1032 Siderite, 948 Siderophore, 978–79 Sigma (s) bonds, 351–52, 354 Sigma (s) molecular orbitals, 359 Sigma (s) symbol, 186 Significant figures, 22–25, 1054 in calculations, 23–25 Silica, reaction with hydrofluoric acid, 929 Silica glass (quartz glass), 952 Silicates, 950–52 Silicon (Si), 8, 268, 476, 487, 950 doping of, 489–90 electronic properties of, 488 Lewis symbol for, 290 nonbonding electron pairs in, 477 occurrence and preparation of, 950 semiconductor, 268 surface of, 41 thermodynamic quantities for, 1061 Silicon carbide, 486, 510, 949 Silicon dioxide, 917, 919, 950 Silicones, 952 Silicon tetrachloride, 72 Silver (Ag), 8, 51, 296 alloys of, 474, 476 corrosion of, 131 mole relationships of, 88 on nanoscale, 498 oxidation of, 135–37 reaction with copper, 135–36 as reducing agent, 846 sterling, 513 thermodynamic quantities for, 1061 Silver chloride, 184, 648 Silver chromate, 724 Silver ion (Ag+), 60 Simple cubic (primitive cubic), 467 Single bonds, 298, 919 bond enthalpies of, 316 length of, 318–20 rotations about, 1009 Single-walled carbon nanotubes, 499 SiO2, 951, 952 SI units, 15–17 base units, 15 density, 19 derived, 18 length and mass, 15, 16–17 for speed, 18 temperature, 15, 17–18 volume, 18–19 Skeleton, carbon-carbon, 1007 Skyscrapers, 784 Slaked lime, 199 Slater, John, 253 Slater’s rules, 253 Smalley, Richard, 498 Smectic liquid crystalline phases, 449 Smog, 592, 760–61 Soap, 127n, 652, 1007, 1026–27 Soap scum, 770
Society of Automotive Engineers (SAE), 437 Soda-lime glass, 952 Sodium (Na), 8, 49, 242, 469 condensed electron configuration of, 252 cubic structure of, 506 effective nuclear charge of, 252 electron configuration of, 230, 231 ions of, 54 Lewis symbol for, 290 oxidation in aqueous solution, 136 properties of, 269 reactions of with chlorine, 291–92 with oxygen, 270 in seawater, 765 Sodium–24, 893 Sodium acetate, 519–20, 704 Sodium azide, 83, 398–99 Sodium bicarbonate (baking soda), 124, 129–30 standard enthalpy of formation for, 184 Sodium borohydride, 953 Sodium bromide, 928 Sodium carbonate, standard enthalpy of formation for, 184 Sodium cation, 54 Sodium chloride, 56 conductivity of solution of, 116, 117 coordination environments in, 483 crystal structure of, 291 dissolution in water, 515–16 electrolysis of aqueous, 860 molten, 860 formation of, 291–92, 295 standard enthalpy of formation for, 184 states of, 427 structure of, 482–83 Sodium fluoride, 483, 730 Sodium formate, 746 Sodium hydroxide, 6, 127–28, 652, 1027 Sodium hypochlorite, 678, 927 Sodium ion (Na+), 60 Sodium ion batteries, 284 Sodium lactate, 709 Sodium monofluorophosphate, 730 Sodium perchlorate, 915 Sodium propionate, 1026 Sodium silicate, 109 Sodium stearate, 543–44 Sodium sulfate, dissociation of, 118 Sodium tripolyphosphate, 943, 975 Sodium vapor lights, 213, 270 Softening of water, 770 Solar cells (photovoltaic devices), 5, 193 Solar energy, 191–92, 370 advances in, 20–21 Solar spectrum, 752 Solder, plumber, 473 Solid(s), 7 amorphous, 465 classifications of, 464 concentration of, 623 covalent-network, 464, 486–90 crystalline, 427, 465 intermolecular attractive forces in, 426 ionic, 464, 481–85
empirical formula and density of, 484–85 properties of, 481 structures of, 482–85 in water, 516 metallic. See Metal(s) molecular, 486 molecular comparison of liquids and, 426–28 polymeric, 490–96 properties of, 426 structures of, 465–76 close packing of spheres, 470–71 unit cells, 465–67 Solid solutes, 525 Solid solutions, 474, 513 Solubility, 120, 518–20, 726–34 amphoterism and, 733–34 common-ion effect and, 726–27 complex ion formation and, 731–33 molar, 723 of organic substances, 1007 pH and, 728–30 pressure effects on, 523–25 solubility-product constant and, 722–26 solubility-product constant vs., 723–26 solute-solvent interactions and, 520–23 temperature effects on, 525–26 Solubility equilibria, 722–26 solubility-product constant, 723, 1063 limitations of, 726 reaction quotient and, 734 solubility vs., 723–26 Solubility guidelines, for ionic compounds, 120–21 Solubility-product constant (Ksp, solubility product), 723, 1063 limitations of, 726 Solutes, 116, 513 molarity to calculate grams of, 141 solid, 525 titration to determine quantity of, 145–46 Solute–solute interactions, 515 Solute-solvent interactions, 520–23 Solution(s), 10, 11, 116, 512–55 acidic, 683–85 aqueous. See Aqueous solution(s) basic, 683–85 balancing equations for reactions in, 833–35 buffered, 707–13 blood as, 707, 713 buffer capacity and pH, 710–11 calculating pH of buffer, 708–10 composition and action of, 707–8 strong acids or bases in, 711–13 colligative properties, 530–41 boiling-point elevation, 533–34 of electrolyte solutions, 540–41 freezing-point depression, 534–36 molar mass determination through, 539–40 osmosis, 536–39 vapor pressure reduction, 530–33 colloids, 541–46 hydrophilic and hydrophobic, 542–44 removal of colloidal particles, 544–46
Index types of, 541 concentration of, 139–43, 526–30 conversion of units of, 528–30 dilution, 141–43 of electrolyte, 140 interconverting molarity, moles, and volume, 140–41 in mass percentage, 526–27 molality, 527–29 molarity, 139–40, 527–30 in mole fraction, 527–29 in parts per billion (ppb), 526–27 in parts per million (ppm), 526–27 defined, 116 examples of, 513 formation of, 514–18 energetics in, 515–17 intermolecular forces and, 514–15 reactions and, 517–18 spontaneity, entropy, and, 515 hypertonic, 537 hypotonic, 537 ideal, 532–33 isotonic, 537 neutral, 659, 683–85 preparing by dilution, 143 saturated, 518–20, 722–23 solid, 474, 513 standard, 145 stock, 141 supersaturated, 519, 520 unsaturated, 519 Solution alloys, 474 Solution stoichiometry, 144–48 titrations, 145–48 acid-base, 714–22 Solvation, 118, 515 Solvent(s), 116, 513 environmentally friendly, 773–75 ethers as, 1024 ketones as, 1025 supercritical fluid, 773 water as, 703 Solvent–solute interactions, 515 Solvent–solvent interactions, 515 s orbitals, 222–24 energy-level diagrams/electron configurations, 365–66 phases in, 363–64 sp2 hybrid orbitals, 348–50, 351 sp3hybrid orbitals, 348–50, 351 Space-filling models, 54, 332 Space shuttle, 929 Spearmint, 1025 Specific heat, 175–77 Spectator ions, 123, 124 Spectrochemical series, 989 Spectroscopy, measuring reaction rates with, 564 Spectrum, 213 continuous, 213 Speed of light, 208, 209, 214 root-mean-square (rms) vs. average, 404 “Speed bumps” features, 30 Spheres, close packing of, 470–71 sp hybrid orbitals, 346–48, 351 Spinel, 511 Spin magnetic quantum number (ms), 227
Spin-pairing energy, 990–91 Spontaneous processes, 175, 786–90 criterion for, 788 exothermic processes and, 516–17 free energy and, 803–4 oxidation-reduction reactions, 835, 845–46, 847 pressure and, 786–87 reversible and irreversible, 788–90 solution formation and, 514, 515 temperature and, 786–87 Sport drinks, 143 Square lattice, 465, 466 Square-planar complexes, 991–95 Square planar geometry, 334, 340, 341, 343, 973, 977 Square pyramidal geometry, 340, 341 Stability(ies) belt of, 880–81 nuclear even vs. odd number of nucleons, 882–83 magic numbers and, 882 neutron-to-proton ratio, 880–82 radioactive series (nuclear disintegration series), 882 of organic substances, 1007 Stack, fuel cell, 857 Stained glass, 498, 962, 963 Stainless steel, 473, 474, 858 Stalactites and stalagmites, 203 Standard atmospheric pressure, 386 Standard cell potential, 838–45 Standard deviation, 22, 1057 Standard emf, 838–45 Standard enthalpy change, 183 Standard enthalpy of formation, 184 Standard free-energy change, 806 Standard free energy of formation, 806 Standard hydrogen electrode (SHE), 839–40 Standard molar entropies, 801–2 Standard pressure, 801n Standard reduction (half-cell) potentials, 839–43, 1064 Standard solution, 145 Standard state, 183 Standard temperature and pressure (STP), 392 Standing waves, 219 Stannous fluoride, 730 Stannous or tin(II) ion (Sn2+), 60 Star, formation of, 903 Starch, 188, 490, 1036–37 Starfish, 728 State(s) changes of, 12 of gas, 387 of matter, 7 of reactants and products, 81 State function(s), 167–69, 788 enthalpy as, 169 Static equilibrium, 611 Statistical thermodynamics, 795 Steam turbines, 824 Stearate, 1007 Stearic acid, 550 Steel, 474 mild, 474 stainless, 473, 474, 858 Stereoisomers, 981, 982–85
Sterling silver, 473, 513 Stern, Otto, 227 Sternbach, Leo, 330 Stern-Gerlach experiment, 227 STM (scanning tunneling microscopy), 41 Stock solutions, 141 Stoichiometrically equivalent quantities, 96 Stoichiometry, 77–113 Avogadro’s number and the mole, 86–92 interconverting masses and moles, 90–91 interconverting masses and numbers of particles, 91–92 molar mass, 88–89 calculation, 711, 712 chemical equations, 78–81 balancing, 78–80 states of reactants and products, 81 defined, 78 empirical formulas from analyses, 92–96 combustion analysis, 95–96 molecular formula from, 94 formula weights, 84–86 percentage composition from formulas, 85–86 of half-reaction, 861 limiting reactions (limiting reagent), 99–103 theoretical yields, 102–3 patterns of chemical reactivity, 81–84 combination and decomposition reactions, 81–83 combustion reactions, 83–84 problem-solving procedure for, 144 quantitative information from balanced equations, 96–99 reaction rates and, 562–63 solution, 144–48 titrations, 145–48 Stony corals, 703–4, 728 STP (standard temperature and pressure), 392 Straight-chain alkanes, 1009 Straight-chain hydrocarbons, 1009 Straight-run gasoline, 1014 Stratosphere, 750 Strong acids, 125–26, 656–57, 664–66 added to buffers, 711–13 Strong acid-strong base titrations, 714–16 Strong bases, 125–26, 665–66 added to buffers, 711–13 Strong electrolytes, 118–19, 664 identifying, 126–27 Strong-field ligands, 989 Strong nuclear forces, 46 Strontium (Sr), 272 electron configuration of, 233 in seawater, 765 Strontium–90, 886, 890–91, 911 Strontium ion (Sr2+), 60 Strontium oxide, 284, 821 Structural formulas, 54 condensed, 1009, 1011, 1012–13 Structural isomers, 67, 981 of alkanes, 1009–10 of butene, 1015 Structure, atomic. See Atomic structure; Electronic structure
I-21
Structure, chemical, 4 acid-base equilibria and, 685–89 binary acids, 685–86 carboxylic acids, 688–89 factors affecting acid strength, 685 oxyacids, 686–88 Styrene, 108, 772 Subatomic particles, 41 Subcritical mass, 897 Sublimation, 438–39 heat of, 439 Sublimation curve, 446 Submicroscopic realm, 5 Subscript coefficient vs., 79 in formulas, 84 Subshell, 221 Substance, 7, 11 SI unit for amount of, 15 Substitutional alloys, 474, 475, 476 Substitution reactions, 1020–21 Substrates, 593 Subtraction in exponential notation, 1052 significant figures and, 24 Sucrose, 1035–36 conductivity of solution of, 116, 117, 118 dehydration of, 936 properties of, 486 reactions in dilute acid solutions, 602 reaction with sulfuric acid, 936 standard enthalpy of formation for, 184 van’t Hoff factor for, 541 Sugar, 1035–36 density of, 19 invert, 1036 Sugarcane, 158 bioethanol from, 192 Sulfate ion, 62, 63, 936 Sulfates, 936 in seawater, 765 Sulfide ions, 63, 736 Sulfides, 269, 935 acid-insoluble, 738 base-insoluble, 738 Sulfites, 935 Sulfur (S), 8, 267, 274, 476, 934–35 electron configuration of, 249 elemental, 274, 371 Lewis symbol for, 290 nonbonding electron pairs in, 477 occurrences and production of, 934 oxides, oxyacids, and oxyanions of, 935–36 properties and uses of, 273, 934–35 Sulfur compounds, in troposphere, 758–60 Sulfur dioxide, 274, 384, 758–60, 821, 932, 935 in atmosphere, 751, 758–60 dissolved in water, 932 reaction with calcium carbonate, 759 Sulfuric acid, 125, 674, 686, 758, 931, 935–36 acid-dissociation constant of, 674 commercial, 936 formula weight of, 85 reaction with sucrose, 936 sale of, 6 Sulfurous acid, 674
I-22
Index
Sulfur oxides, 934 Sulfur tetrafluoride, 380 Sulfur trioxide, 936 Sun, as energy source, 875 Superconductors, 475 Supercooling, 441 Supercritical fluid, 442, 445, 446 Supercritical fluid extraction, 442 Supercritical fluid solvents, 773 Supercritical mass, 897 Supercritical water, 773 Superhydrophobic surfaces, 425, 426 Supernova explosion, 903 Superoxide ion, 270, 274, 378 Superoxides, 933 Supersaturated solutions, 519, 520 Surface tension, 437 Surfactants, 1007 Surroundings, 162–63 entropy changes in, 802–3 Sustainability, 158 Symbols, chemical, 8 Synchrotron, 884 Syndiotactic polypropylene, 511 Synthetic diamond, 946 Synthetic Genomics, 20 Syringes, 18, 19 System, 162–63 Système International d’Unités. See SI units Systolic pressure, 388 Szilard, Leo, 898 Table salt. See Sodium chloride Talc, 950 Tartaric acid, 156, 697 acid-dissociation constant of, 674 Taste, 651 Tausonite, 507 Technetium, 250 Technetium–99, 893, 913 Teflon (polytetrafluoroethylene), 510, 773, 927 Television, 208 Tellurium (Te), 273, 934–35 Temperature absolute, 403, 578 of atmosphere, 750 body, 180 color as function of, 210 critical, 441–42 Curie, 968, 1000 of Earth’s surface, 761–62 entropy and, 800–801 equilibria and, 634–37 fusion and, 902 Gibbs free energy and, 809–10 kinetic energy and, 578 Le Châtelier’s principles and, 631 molecular speeds and, 403–4 Néel, 968 pressure and, 1056 reaction rates and, 558, 575–81 activation energy, 577–78, 579–81 Arrhenius equation for, 578–79 collision model of, 576 orientation factor in, 576, 577 real vs. ideal gas behavior and, 410, 411 regulation in humans, 180 of seawater, 766 climate change and, 781 SI units of, 15, 17–18
solubility and, 525–26 spontaneity of reaction and, 787, 810 spontaneous processes and, 786–87 standard enthalpy change, equilibrium constant, and, 811 vapor pressure and, 443–44 volume and, 389–90 Tentative explanation (hypothesis), 15 Tera prefix, 16 Terephthalic acid, 773 Termolecular reaction, 581 Tertiary structure, of proteins, 1033 Test-taking strategies, 103 Tetraboric acid, 953 Tetracene, 381 2,2,3,3-Tetrachlorobutane, 1018 Tetraethyl lead, 1014 Tetragonal lattice, 466 Tetrahedral complexes, 991–95 Tetrahedral geometry, 332, 335, 337, 343, 351, 973, 1006 Tetrahydrofuran (THF), 1024 tetra- prefix, 65 Thallium, electron configuration of, 233 Thallium–201, 893 Theoretical yields, 102–3 Theory, 15 Thermal conductivity, 469, 478 Thermal energy, 162 Thermal pollution, 526 Thermite reaction, 168, 201, 912 Thermochemical equations, 173 Thermochemistry, 158–205 calorimetry, 175–79 bomb (constant-volume), 178–79 constant-pressure, 177–78 heat capacity and specific heat, 175–77 defined, 160 energy and, 160–64 fossil fuels, 190–91 kinetic and potential, 160–62 nuclear, 191 solar, 191 system and surroundings, 162–63 transferring, 163–64 units of, 162 enthalpy(ies), 169–75 defined, 169 of formation, 183–87 of reaction, 172–74, 185–87 spontaneous processes and, 175 first law of thermodynamics, 164–69 algebraic expression of, 166 endothermic and exothermic processes, 167, 168 heat and work related to internal energy changes, 165–67 internal energy, 164–65 state functions, 167–69 of foods, 188–90 of fuels, 190–91 Hess’s law, 181–83 Thermodynamic equilibrium constant, 618 Thermodynamics, 160, 784–825 defined, 786 entropy, 790–803 absolute, 800, 801, 802 of expansion, 792 heat transfer and temperature related to, 790
life and, 800 microstates and, 794–96 molecular interpretation of, 793–800 probability and, 795 in reactions, 800–803 temperature and, 800–801 first law of, 164–69 algebraic expression of, 166 endothermic and exothermic processes, 167, 168 heat and work related to internal energy changes, 165–67 internal energy, 164–65 state functions, 167–69 Gibbs free energy, 803–8 equilibrium constant and, 811–15 under nonstandard conditions, 811–12 spontaneity and, 803–4 standard free-energy changes, 806 temperature and, 811 second law of, 792–93 spontaneous processes, 786–90 pressure and, 786–87 reversible and irreversible, 788–90 temperature and, 786–87 statistical, 795 third law of, 799 Thermodynamic sign convention, 263n Thermonuclear reactions, 902 Thermoplastics, 492 Thermosetting plastics, 492 Thermosphere, 750 THF (tetrahydrofuran), 1024 Thiosulfate ion, 936 Third law of thermodynamics, 799 THMs (trihalomethanes), 770–71 Thomson, J. J., 42, 43 Thomson, William (Lord Kelvin), 390 Thorium–232, 904 Thorium–234, 877 Threonine, 1031 Thymine, 459, 1040, 1041 Thymol blue, 664 Thyroxin, 928 Tiling, of unit cells, 465, 466 Time reaction rates and, 559, 561, 569–75 SI unit of, 15 Tin (Sn), 8, 265, 950 bonding between chlorine and, 304 gray, 510 oxidation in aqueous solution, 136 white, 510 Tin(II) or stannous ion (Sn2+), 60 Tire gauges, 387 Titan (moon), 401 Titanic, raising the, 873 Titanium dioxide, 370 Titanium tetrachloride, 867 Titration(s), 145–48 acid-base, 714–22 of polyprotic acids, 720–21 strong, 714–16 weak, 716–19 end point of, 146n equivalence point of, 145 TNT (trinitrotoluene), 319, 940 Tokamak, 902
Toluene (methylbenzene), 108, 486, 532, 551, 773, 1019 Tooth decay, 722, 730 Tooth enamel, 722, 746 Torr, 386, 388 Torricelli, Evangelista, 385–86 Trace elements, 58 trans fats, 1037 Transferrin, 978, 979 trans isomers, 969–70, 982 Transition-metal complexes, 968–74 Transition-metal ions, 296 aqueous solutions of, 966 Transition-metal oxides, 592 Transition metals, 59, 964–68 chromium, 968 compounds of. See Coordination compounds copper. See Copper (Cu) electron configurations of, 232, 965–67 iron. See Iron (Fe) magnetism, 967–68 paramagnetism, 967 mineral sources of, 964 oxidation states of, 965–67 physical properties of, 964–65 position in periodic table, 964 radii of, 964–65 Transition state (activated complex), 577, 582 Translational motion, 796 Transplatin, 379 Transuranium elements, 52, 885–86 Triazine, 327 Triclinic lattice, 466 Trifluoromethyl acetylacetonate (tfac), 1001 Trigonal bipyramidal geometry, 333, 335, 339, 340, 343, 1001 Trigonal planar geometry, 333, 334, 335, 337, 343, 351, 1006 Trigonal pyramidal geometry, 333, 334, 336, 337 Trihalomethanes (THMs), 770–71 Trimethylamine, 1028 2,2,4-Trimethylpentane, 1014 2,3,4-Trimethylpentane, 201 Trinitroglycerin, 193–94 Trinitrotoluene (TNT), 319, 940 Trinity test, 898 Triple-alpha process, 903 Triple bonds, 298, 1006 hybrid orbitals and, 353–54, 358 length of, 318–20 Triple point, 445, 446 tri- prefix, 65 Tristearin, 188 Tritium, 74, 921 Tropopause, 750 Troposphere, 750 sulfur compounds and acid rain, 758–60 Trouton’s rule, 823 Tryptophan, 1031 T-shaped geometry, 333, 340 Tumor, malignant, 907 Tums, 130 Tungsten, 280 Tungsten carbide, 949 Turnover number, 593 20-kiloton bomb, 898
Index Twisted nematic liquid crystal display, 451 2p orbitals, molecular orbitals from, 362–65 Tyndall effect, 541–42 Tyrosine, 1031 Uhlenbeck, George, 227 Ultraviolet photoelectron spectroscopy (UPS). See Photoelectron spectroscopy (PES) Ultraviolet radiation, 208 Uncertainty principle, 217–19 measurement and, 218 UNICEF, 945 Unimolecular reactions, 581 Unit cells, 465–67 United Nations Children’s Fund (UNICEF), 945 Universe, entropy of, 793 Unpaired electrons, 229 Unsaturated hydrocarbons, 1015 alkenes, 1015–17 alkynes, 1017–19 aromatic, 1008, 1019–21 Unsaturated solutions, 519 Uracil, 1040 Uranium (U), 233 isotopes of, 408, 876 Uranium, isotopes of, 247 Uranium–233, 896 Uranium–234, 876 Uranium–235, 408, 876, 896–98 Uranium–238,408, 876, 882, 886, 888, 894, 900–901 abundance of, 904 rate of decay, 886 Urbain, G., 246 Urea, 188, 606, 1006 Urease, 605 Uric acid, 271 Valence band, 487 Valence-bond theory, 345 hybrid orbitals and, 346–51 involving d orbitals, 349–50 pi bonding and, 352–58 sp, 346–48 sp2 and sp3, 348–50 VSEPR model and, 345, 346, 350 Valence electron configurations, 233–35, 236, 237 Valence electrons, 231, 233–35, 236, 237 bonding and, 290 delocalized, 469 effective nuclear charge of, 252, 254 less than an octet of, 312–13 more than an octet of, 313–14 Valence orbitals, 249 Valence-shell electron-pair repulsion (VSEPR) model, 334–43 basis of, 335–36 for larger molecules, 342–43 for molecules with expanded valence shells, 339–41 nonbonding electrons and multiple bonds, 338–39 valence-bond theory and, 345, 346, 350 Valence shells, molecules with expanded, 339–41 Valine, 1031 Valium (Diazepam), 330, 331–32
Valproic acid, 109 Vanadium, 474 Vancomycin, 21 van der Waals, Johannes, 411, 428 van der Waals constants, 412 van der Waals equation, 411–13 van der Waals forces comparison of, 435 dipole-dipole, 428 hydrogen bonding, 431–34 trends in, 431–32 in water, 432, 433–34 ion-dipole, 428 London dispersion, 429–30 van der Waals radius (nonbonding atomic radius), 254 Vanilla, 1025 Vanillin, 112 van’t Hoff factor (i), 540–41 Vapor, 7, 384. See also Gas(es) Vaporization enthalpies of, 183 heat of, 439 Vapor pressure, 442–45 boiling point and, 444 lowering, 530–33 molecular-level explanation of, 443 volatility, temperature, and, 443–44 Vapor-pressure curve, 446 Variables calculations involving many, 393 dependent, 1056 independent, 1056 Vector quantities, 343 Vectors, 465n Vehicles, flex-fuel, 202 Vibrational motion, 796 Vinegar, 1025 as household acid, 124 Vinyl alcohol, 342 Vinyl chloride, 927 Viscosity, 437 Visible light, 208, 209 color and, 986 Visible spectrum, 986 Vision, chemistry of, 357 “Visualizing Concepts” feature, 30 Vitamins A (retinol), 522 B, 522 B6, 548 C (ascorbic acid), 127n, 522, 551, 674, 1007, 1025 D, 522 E, 522, 548 fat- and water-soluble, 522 K, 522 VO2 max, 418 Volatile, 443 Volatile components, separating, 532 Volatile organic compounds, 773 Volatile substance, 530 Volatility, 443 Volcanic glass (obsidian), 465 Volcanoes, 757 Volta, Alessandro, 853 Voltaic (galvanic) cells, 835–37. See also Batteries concentration cells, 852–54 electromotive force (emf) in, 838–45 concentration effects on, 849–54
equilibrium and, 850 oxidizing and reducing agents, 843–45 standard reduction (half-cell) potentials, 839–43 molecular view of electrode process, 865 standard cell potential of, 839, 842 work done by, 849 Volume(s), 169 conversions involving, 28–29 equilibria and, 633–34 of gas, 397–99 interconverting molarity, moles, and, 140–41 law of combining, 390 molar, 392 pressure-volume relationship, 388–89 quantity-volume relationship, 390–91 real vs. ideal gas behavior and, 410 SI unit of, 18–19 temperature-volume relationship, 389–90 Volumetric flasks, 18, 19 von Hevesy, G., 246 Vulcanization, 496, 934 Waage, Peter, 614 Washing soda, 109, 947 Wastes nuclear, 900–901 oxygen-demanding, 768 Watches, 451 Water, 9, 53. See also Aqueous equilibria; Aqueous solution(s) acidity of, 687 as analogy for electron flow, 838 arsenic in drinking, 156, 526, 945 autoionization of, 658–60 boiling-point elevation of, 534 bonding in, 338 hydrogen bonding, 432, 433–34 chlorine dissolved in, 770–71 collecting gases over, 401–2 critical temperature and pressure of, 442 density of, 19 desalination of, 768–69 dissolution in, 116, 117 of ionic solid, 798 of oxygen, 768 of sodium chloride, 515–16 of Earth, 764–71 freshwater and groundwater, 766–67 global water cycle, 764 human activities and, 767–71 salt water, 765–66 seawater, 728 electrolysis of, 9 evaporation of, 12 excess consumption of, 143 forms of, 384 fresh, 766–67 H+ ions in, 652–53 hard, 156, 770 heating curve for, 440–41 heavy, 920, 921 of hydration, 518 ionic compounds in, 117–18 ion product of, 659–60, 679
I-23
meniscus, 438 metal ions in, 682 molar mass of, 88 molecular compounds in, 118 molecular model of, 4 from oxidation of glucose, 98 perchlorate in drinking, 930 phase changes of, 764 phase diagram of, 446–48 photodissociation of, 782 physical states of, 7 polarity of, 343 properties of, 9, 1058 reactions of, 115 with alkali metals, 269 with ammonia, 655 with anions, 681–82 with barium oxide, 932–33 with butyl chloride, 561 with calcium, 272 with calcium hydride, 923 with carbon dioxide, 267–68 with cations, 682–83 with chlorine, 275 with hydrogen chloride, 652 with nitrous acid, 655 softening of, 770 solubility of guidelines for ionic compounds, 120–21 solubility of gases in, 520, 524, 525–26 as solvent, 117–18, 625, 703 specific heat of, 176 standard enthalpy of formation for, 184 standard reduction potentials in, 841 structural formula for, 54 supercritical, 773 surface tension of, 437 treatment of municipal supplies, 769–71 valence bonding in, 349 vapor pressure of, 444, 531 vibrational and rotational motion in, 796 wave characteristics of, 208 Water gas, 922 Water purification, 769–71 Water softening, 770 Water-soluble vitamins, 522 Water vapor climate and, 761–64 standard enthalpy of formation for, 184 Watt (W), 197 Wave behavior of matter, 216–19 Wave functions, 219 Wavelength, 208–9 frequency calculated from, 210 Wave mechanics, 219 Wave nature of light, 208–10 Waves electrons as, 217 nodes on, 219 standing, 219 Weak acid(s), 125–26, 656–57, 666–75 acid-dissociation constant, 667–68, 670–73, 679–81 common-ion effect on, 704–7 percent ionization of, 669 polyprotic acids, 674–75
I-24
Index
Weak acid-strong base titrations, 716–19 Weak base(s), 125–26, 676–79 common-ion effect on, 704–7 types of, 677–79 Weak electrolytes, 118–19 identifying, 126–27 Weak-field ligands, 989 Weak nuclear force, 46 Weather air densities and, 396 gases and, 382–83 Weather balloon, 388 WebElements (Web site), 30 Weight atomic, 47–49, 85, 251 density vs., 19 mass vs., 16n molecular, 71 Welding, 931 Werner, Alfred, 969–70, 1002
Werner’s theory, 969–71 Wet chemical methods, 736 “What’s Ahead” features, 30 White dwarfs, 903 White light, 986 White phosphorus, 942 Wind energy, 191 Wires, quantum, 498 Wöhler, Friedrich, 1006 Wood, fuel value and composition of, 190 Wood metal, 473 Work, 160 electrical, 172 internal energy change and, 165–67 mechanical, 172 pressure-volume (mechanical work), 170–71, 172 sign conventions for, 166 transferring energy and, 163–64 Work function, 212
World Health Organization, 771 World ocean, 765–66
Young Girl Reading (painting), 5
Xenon, 276, 751 Xenon compounds, 925 Xenon hexafluoride, 379 Xenon tetrafluoride, 925 X-ray crystallography, 468 X-ray diffraction, 217, 468 X-ray diffractometers, 468 X-ray photoelectron spectroscopy (XPS), 286 X-rays, 208, 209, 212, 228 para-Xylene, 773
Zepto prefix, 16
Yucca Mountain, 901
Zeros, significant digits and, 23 Zinc (Zn), 232 in cathodic protection, 859 in galvanized iron, 859 oxidation of, 135, 136 reaction with hydrochloric acid, 828 as reducing agent, 844 in solution of Cu2+, 835, 838 Zinc blende, 482–83, 486 Zinc ion, 60, 285
Yellow brass, 473 Yield actual, 102–3 percent, 102–3 theoretical, 102–3
Zinc sulfide, 892 Zinn, Walter, 898 Zirconium, 246, 965 Zone refining, 950
Common Ions Positive Ions (Cations) 1ⴙ ammonium (NH 4 +) cesium (Cs +) copper(I) or cuprous (Cu+) hydrogen (H +) lithium (Li +) potassium (K +) silver (Ag +) sodium (Na+)
mercury(II) or mercuric (Hg 2+) strontium (Sr 2+) nickel(II) (Ni 2+) tin(II) or stannous (Sn2+) zinc (Zn2+)
2ⴙ barium (Ba2+) cadmium (Cd2+) calcium (Ca2+) chromium(II) or chromous (Cr 2+) cobalt(II) or cobaltous (Co2+) copper(II) or cupric (Cu2+) iron(II) or ferrous (Fe 2+) lead(II) or plumbous (Pb 2+) magnesium (Mg 2+) manganese(II) or manganous (Mn2+) mercury(I) or mercurous (Hg2 2+)
Negative Ions (Anions) 1ⴚ acetate (CH 3COO - or C2H 3O2 -) bromide (Br -) chlorate (ClO3 -) chloride (Cl -) cyanide (CN -) dihydrogen phosphate (H 2PO4 -) fluoride (F -) hydride (H -) hydrogen carbonate or bicarbonate (HCO3 -)
3ⴙ aluminum (Al3+) chromium(III) or chromic (Cr 3+) iron(III) or ferric (Fe 3+)
hydrogen sulfite or bisulfite (HSO3 -) hydroxide (OH -) iodide (I -) nitrate (NO3 -) nitrite (NO2 -) perchlorate (ClO4 -) permanganate (MnO4 -) thiocyanate (SCN -) 2ⴚ carbonate (CO3 2-) chromate (CrO4 2-) dichromate (Cr2O7 2-) hydrogen phosphate (HPO4 2-) oxide (O 2-) peroxide (O2 2-) sulfate (SO4 2-) sulfide (S 2-) sulfite (SO3 2-) 3ⴚ arsenate (AsO4 3-) phosphate (PO 4 3-)
Fundamental Constants* Atomic mass unit Avogadro’s number Boltzmann’s constant Electron charge Faraday’s constant Gas constant
1 amu 1g NA k e F R
Mass of electron
me
Mass of neutron
mn
Mass of proton
mp
Pi Planck’s constant Speed of light in vacuum
p h c
= = = = = = = = = = = = = = = = =
1.660538782 * 10-27 kg 6.02214179 * 1023 amu 6.02214179 * 1023>mol 1.3806504 * 10-23 J>K 1.602176487 * 10-19 C 9.64853399 * 104 C>mol 0.082058205 L-atm>mol-K 8.314472 J>mol-K 5.48579909 * 10-4 amu 9.10938215 * 10-31 kg 1.008664916 amu 1.674927211 * 10-27 kg 1.007276467 amu 1.672621637 * 10-27 kg 3.1415927 6.62606896 * 10-34 J-s 2.99792458 * 108 m>s
*Fundamental constants are listed at the National Institute of Standards and Technology Web site: http://www.nist.gov/physlab/data/physicalconst.cfm
Useful Conversion Factors and Relationships Length
Energy (derived)
SI unit: meter (m)
SI unit: Joule (J) 1 J = 1 kg-m2>s2
1 km = 0.62137 mi 1 mi = 5280 ft
= 0.2390 cal
= 1.6093 km
= 1 C-V
1 m = 1.0936 yd
1 cal = 4.184 J
1 in. = 2.54 cm (exactly)
1 eV = 1.602 * 10-19 J
1 cm = 0.39370 in. 1 Å = 10-10 m
Pressure (derived) SI unit: Pascal (Pa)
Mass
1 Pa = 1 N>m2 = 1 kg>m-s2
SI unit: kilogram (kg)
1 atm = 1.01325 * 105 Pa
1 kg = 2.2046 lb 1 lb = 453.59 g
= 760 torr = 14.70 lb>in2
= 16 oz 1 amu = 1.660538782 * 10-24 g
1 bar = 105 Pa 1 torr = 1 mm Hg
Temperature Volume (derived)
SI unit: Kelvin (K)
SI unit: cubic meter (m3)
0 K = -273.15 °C
1 L = 10-3 m3
= -459.67 °F
= 1 dm3
K = °C + 273.15 °C = °F =
5 9 9 5
= 103 cm3
(°F - 32°)
= 1.0567 qt
°C + 32°
1 gal = 4 qt = 3.7854 L 1 cm3 = 1 mL 1 in3 = 16.4 cm3
Color Chart for Common Elements
Generic metal
Ag Silver
Au Gold
Br Bromine
C Carbon
Ca Calcium
Cl Chlorine
Cu Copper
F Fluorine
H Hydrogen
I Iodine
K Potassium
Mg Magnesium
N Nitrogen
Na Sodium
O Oxygen
P Phosphorus
S Sulfur
Si Silicon
Periodic Table of the Elements Main Group Representative Elements
1
2
3
4
5
6
7
1Aa 1 1 H 1.00794
2A 2
3 Li
4 Be
6.941
9.012182
11 Na
12 Mg
22.989770
24.3050
19 K
Main Group Representative Elements
5 B
4A 14 6 C
5A 15 7 N
10.811
12.0107
14.0067
13 Al
14 Si
15 P
3A 13 Metals
Metalloids
Nonmetals
Transition metals 8B 9 27 Co
20 Ca
3B 3 21 Sc
4B 4 22 Ti
5B 5 23 V
39.0983
40.078
44.955910
47.867
50.9415
37 Rb
38 Sr
39 Y
40 Zr
41 Nb
42 Mo
43 Tc
44 Ru
45 Rh
46 Pd
85.4678
87.62
88.90585
91.224
92.90638
95.94
[98]
101.07
102.90550
55 Cs
56 Ba
71 Lu
72 Hf
73 Ta
74 W
75 Re
76 Os
77 Ir
132.90545
137.327
174.967
178.49
180.9479
183.84
186.207
190.23
192.217
6B 6 24 Cr
7B 7 25 Mn
8 26 Fe
51.9961 54.938049 55.845
1B 11 29 Cu
2B 12 30 Zn
63.546
47 Ag
106.42
78 Pt
10 28 Ni
58.933200 58.6934
8A 18 2 He
6A 16
7A 17
4.002602
8 O
9 F
10 Ne
15.9994 18.998403 20.1797
16 S
17 Cl
18 Ar
32.065
35.453
39.948
31 Ga
32 Ge
33 As
34 Se
35 Br
36 Kr
65.39
69.723
72.64
74.92160
78.96
79.904
83.80
48 Cd
49 In
50 Sn
51 Sb
52 Te
53 I
54 Xe
107.8682
112.411
114.818
118.710
121.760
127.60
79 Au
80 Hg
81 Tl
82 Pb
83 Bi
84 Po
85 At
86 Rn
26.981538 28.0855 30.973761
126.90447 131.293
200.59
204.3833
207.2
208.98038
[208.98]
[209.99]
[222.02]
87 Fr
88 Ra
103 Lr
104 Rf
105 Db
106 Sg
107 Bh
108 Hs
109 Mt
110 Ds
111 Rg
112 Cn
113
114
115
116
117 **
118
[223.02]
[226.03]
[262.11]
[261.11]
[262.11]
[266.12]
[264.12]
[269.13]
[268.14]
[281.15]
[272.15]
[285]
[284]
[289]
[288]
[292]
[294]
[294]
57 La
58 Ce
59 Pr
60 Nd
61 Pm
62 Sm
63 Eu
64 Gd
65 Tb
66 Dy
67 Ho
68 Er
69 Tm
70 Yb
144.24
[145]
150.36
151.964
157.25
158.92534
162.50
94 Pu
95 Am
96 Cm
97 Bk
98 Cf
99 Es
100 Fm
101 Md
102 No
[244.06]
[243.06]
[247.07]
[247.07]
[251.08]
[252.08]
[257.10]
[258.10]
[259.10]
Lanthanide series
138.9055
Actinide series
89 Ac [227.03]
a The
140.116 140.90765
90 Th
91 Pa
92 U
93 Np
232.0381 231.03588 238.02891 [237.05]
195.078 196.96655
164.93032 167.259 168.93421
labels on top (1A, 2A, etc.) are common American usage. The labels below these (1, 2, etc.) are those recommended by the International Union of Pure and Applied Chemistry (IUPAC). The names and symbols for elements 113 and above have not yet been decided. Atomic weights in brackets are the names of the longest-lived or most important isotope of radioactive elements. Further information is available at http://www.webelements.com ** Discovered in 2010, element 117 is currently under review by IUPAC.
173.04
List of Elements with Their Symbols and Atomic Weights Element Actinium Aluminum Americium Antimony Argon Arsenic Astatine Barium Berkelium Beryllium Bismuth Bohrium Boron Bromine Cadmium Calcium Californium Carbon Cerium Cesium Chlorine Chromium Cobalt Copernicium Copper Curium Darmstadtium Dubnium Dysprosium Einsteinium Erbium Europium Fermium Fluorine Francium Gadolinium Gallium Germanium Gold a
Symbol
Atomic Number
Ac Al Am Sb Ar As At Ba Bk Be Bi Bh B Br Cd Ca Cf C Ce Cs Cl Cr Co Cn Cu Cm Ds Db Dy Es Er Eu Fm F Fr Gd Ga Ge Au
89 13 95 51 18 33 85 56 97 4 83 107 5 35 48 20 98 6 58 55 17 24 27 112 29 96 110 105 66 99 68 63 100 9 87 64 31 32 79
Atomic Weight
Element
227.03a 26.981538 243.06a 121.760 39.948 74.92160 209.99a 137.327 247.07a 9.012182 208.98038 264.12a 10.811 79.904 112.411 40.078 251.08a 12.0107 140.116 132.90545 35.453 51.9961 58.933200 285 63.546 247.07a 281.15a 262.11a 162.50 252.08a 167.259 151.964 257.10a 18.9984032 223.02a 157.25 69.723 72.64 196.96655
Hafnium Hassium Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton Lanthanum Lawrencium Lead Lithium Lutetium Magnesium Manganese Meitnerium Mendelevium Mercury Molybdenum Neodymium Neon Neptunium Nickel Niobium Nitrogen Nobelium Osmium Oxygen Palladium Phosphorus Platinum Plutonium Polonium Potassium Praseodymium Promethium Protactinium Radium
Mass of longest-lived or most important isotope. The names of elements 113 and above have not yet been decided.
b
Symbol
Atomic Number
Hf Hs He Ho H In I Ir Fe Kr La Lr Pb Li Lu Mg Mn Mt Md Hg Mo Nd Ne Np Ni Nb N No Os O Pd P Pt Pu Po K Pr Pm Pa Ra
72 108 2 67 1 49 53 77 26 36 57 103 82 3 71 12 25 109 101 80 42 60 10 93 28 41 7 102 76 8 46 15 78 94 84 19 59 61 91 88
Atomic Weight
Element
178.49 269.13a 4.002602a 164.93032 1.00794 114.818 126.90447 192.217 55.845 83.80 138.9055 262.11a 207.2 6.941 174.967 24.3050 54.938049 268.14a 258.10a 200.59 95.94 144.24 20.1797 237.05a 58.6934 92.90638 14.0067 259.10a 190.23 15.9994 106.42 30.973761 195.078 244.06a 208.98a 39.0983 140.90765 145a 231.03588 226.03a
Radon Rhenium Rhodium Roentgenium Rubidium Ruthenium Rutherfordium Samarium Scandium Seaborgium Selenium Silicon Silver Sodium Strontium Sulfur Tantalum Technetium Tellurium Terbium Thallium Thorium Thulium Tin Titanium Tungsten Uranium Vanadium Xenon Ytterbium Yttrium Zinc Zirconium *b *b *b *b *b *b
Symbol Rn Re Rh Rg Rb Ru Rf Sm Sc Sg Se Si Ag Na Sr S Ta Tc Te Tb Tl Th Tm Sn Ti W U V Xe Yb Y Zn Zr
Atomic Number 86 75 45 111 37 44 104 62 21 106 34 14 47 11 38 16 73 43 52 65 81 90 69 50 22 74 92 23 54 70 39 30 40 113 114 115 116 117 118
Atomic Weight 222.02a 186.207a 102.90550 272.15a 85.4678 101.07 261.11a 150.36 44.955910 266a 78.96 28.0855 107.8682 22.989770 87.62 32.065 180.9479 98a 127.60 158.92534 204.3833 232.0381 168.93421 118.710 47.867 183.84 238.02891 50.9415 131.293 173.04 88.90585 65.39 91.224 284a 289a 288a 292a 294a 294a